- Check whether each choice describes the entire solid or only selected parts. - Pay attention to the difference between “every” and “several.” - Decide whether the condition must hold at all heights.

1. Cavalieri's principle requires equal heights and equal areas for all corresponding cross sections parallel to the base plane. 2. Only choice B states both requirements completely.

B

- Compare the two area expressions at an arbitrary height. - Distinguish cross-sectional shape from cross-sectional area. - Confirm that both solids use the same height interval.

1. For every \(z\) from \(0\) to \(8\), the two cross-sectional areas are equal. 2. The solids also have equal heights. 3. By Cavalieri's principle, \(V_K=V_L\). The different cross-sectional shapes do not matter because their areas are equal.

Yes. The volumes are equal: \(V_K=V_L\).

- Area has square units; ask what multiplying by the layer thickness does to the units. - After finding the volume of one layer, use the fact that the five layers do not overlap.

1. One layer has volume equal to its area times its thickness: \(12\,\text{cm}^2\cdot1\,\text{cm}=12\,\text{cm}^3\). 2. There are \(5\) nonoverlapping identical layers, so the total volume is \(5\cdot12=60\,\text{cm}^3\).

Multiplying the layer area by its \(1\,\text{cm}\) thickness gives \(12\,\text{cm}^3\) per layer. Five such layers have volume \(60\,\text{cm}^3\).

- If you made a thin layer from each picture, how would the two layers compare? - What stays the same when many such layers are stacked to make each tower?

1. Cross section b) is wider than cross section a) and has the same other side length, so each layer of b) has greater area. 2. Both towers have equal heights. Stacking larger-area layers over the same height makes b)'s volume greater, not equal.

No. Tower b) has larger-area slices throughout the same height, so its volume is greater.

- Compare the rectangular footprints of the two columns. - Imagine stacking layers of the same thickness from the bottom upward. - Does the same amount fit in a shorter stack and a taller stack?

1. Both columns have rectangular slices with area \(4\cdot2=8\,\text{cm}^2\). 2. Column b) is taller, so it contains more layers of that same area. 3. Their volumes are not equal: matching slice areas alone do not guarantee equal volumes unless the heights also match.

No. Column b) is taller and contains more equal-area layers. Equal slice areas are not enough when the heights differ.

- Remember that an area cannot be negative. - Compare the simplified expression with volume formulas you know. - Imagine the top base growing until its size matches the bottom base.

1. Substitute \(B_1=B_2=B\) to get \(V=\frac{h}{3}(B+\sqrt{B^2}+B)\). 2. Since the area \(B\) is positive, \(\sqrt{B^2}=B\). Thus \(V=\frac{h}{3}(3B)=Bh\). 3. This is the prism volume formula. As the frustum's two similar bases become the same size, its sides stop tapering and the limiting shape is a prism, not an ordinary pyramid frustum.

a) \(V=Bh\). b) This is the formula for a prism. As the top and bottom base sizes become equal, the sides no longer taper and the shape becomes a prism.

- What is true about every cross section of a prism that is parallel to its bases? - Check both requirements of Cavalieri's principle before calculating a volume. - After justifying equal volumes, use the right prism as the comparison solid.

1. Every cross section parallel to the bases of either prism is congruent to its base, so corresponding cross sections have equal area \(36\,\text{cm}^2\). 2. The prisms also have the same perpendicular height, \(14\,\text{cm}\). Therefore, Cavalieri's principle gives equal volumes. 3. The right prism has volume \(V=Bh=36\cdot14=504\,\text{cm}^3\), so the oblique prism has the same volume. 4. The \(17\,\text{cm}\) lateral-edge length is not needed because the comparison uses the perpendicular height and corresponding cross-sectional areas.

Corresponding cross sections parallel to the bases all have area \(36\,\text{cm}^2\), and the prisms have the same perpendicular height, so Cavalieri's principle applies. The oblique prism's volume is \(504\,\text{cm}^3\). The \(17\,\text{cm}\) lateral-edge length is not needed.

- Compare one horizontal slice before and after the cards are shifted. - Which measurements of that slice change, if any? - Use the unshifted stack only after establishing the Cavalieri comparison.

1. At every height through either stack, a horizontal cross section is an \(8\,\text{cm}\times8\,\text{cm}\) square with area \(64\,\text{cm}^2\). 2. The shifted and unshifted stacks have the same total height, \(22\,\text{cm}\), and equal corresponding cross-sectional areas. 3. Therefore, Cavalieri's principle gives equal volumes. 4. The unshifted stack is a right rectangular prism, so \(V=64\cdot22=1{,}408\,\text{cm}^3\). The shifted stack has the same volume.

The horizontal cross sections have equal area at every height and the stacks have the same height, so Cavalieri's principle shows that shifting the cards does not change the volume. The volume is \(1{,}408\,\text{cm}^3\).

- Compare the column with a simpler solid having the same cross-sectional areas. - The area of each cross section matters more than its exact shape. - Use the common height.

1. Compare the column with a right prism that is \(10\,\text{cm}\) tall and has constant cross-sectional area \(18\,\text{cm}^2\). 2. The two solids have equal cross-sectional areas at every height, so Cavalieri's principle gives equal volumes. 3. Thus, \(V=18\cdot10=180\,\text{cm}^3\).

The column volume is \(180\,\text{cm}^3\).

- Compare all corresponding cross sections, not just one. - A constant cross-sectional area factor produces the same volume factor when heights are equal. - Apply the factor to the known volume.

1. Because the same area scale factor applies to all corresponding cross sections and the heights are equal, the volume has the same scale factor. 2. Therefore, \(V_B=1.25V_A=1.25\cdot320=400\,\text{cm}^3\).

\(V_B=400\,\text{cm}^3\)

- Transfer the ratio of corresponding cross-sectional areas to the volumes. - Identify which part of the ratio corresponds to the known volume. - The unknown volume should be larger than \(270\,\text{cm}^3\).

1. By Cavalieri's principle, a constant ratio of \(3\) to \(2\) for corresponding cross-sectional areas gives the same ratio of volumes. 2. The given \(270\,\text{cm}^3\) corresponds to the smaller, \(2\)-part volume. 3. Therefore, \(V=270\left(\frac{3}{2}\right)=405\,\text{cm}^3\).

The other solid has volume \(405\,\text{cm}^3\).

- What does Cavalieri's principle need to hold for corresponding slices? - Do the slice areas of either solid change with height? - Compare the prism base with the cylinder's circular base.

1. Both solids have a constant cross-sectional area parallel to their bases. 2. To apply Cavalieri's principle, the prism must have the same section area as the cylinder at every height. 3. The required prism base area is therefore \(B=\pi(3)^2=9\pi\,\text{cm}^2\approx28.27\,\text{cm}^2\).

\(B=9\pi\,\text{cm}^2\approx28.27\,\text{cm}^2\).

- Find the known prism volume first. - Replace the unusual column mentally with a prism having the same cross-sectional area. - Work backward from volume to height.

1. The prism volume is \(V=50\cdot12=600\,\text{cm}^3\). 2. By Cavalieri's principle, the column has the same volume as a prism with base area \(40\,\text{cm}^2\) and the same height as the column. 3. Let the column height be \(h\). Then \(40h=600\), so \(h=15\,\text{cm}\).

The column is \(15\,\text{cm}\) tall.

- Decide whether information about only the bottom of a solid determines the entire solid. - Think of solids that narrow at different rates. - State a condition that applies throughout the full height.

1. The claim is false for arbitrary solids because equal base areas and equal heights do not determine the areas of the other cross sections. 2. The missing condition is that, at every corresponding height, cross sections parallel to the base planes must have equal areas.

The claim is false. The solids must have equal heights and equal areas for corresponding cross sections at every height.

- For a cross section made of separate regions, add their areas. - Compare the total cross-sectional areas of the two solids. - Use a simple comparison solid with the same height and constant cross-sectional area.

1. At every height, the total cross-sectional area of solid \(B\) is \(12\,\text{cm}^2+23\,\text{cm}^2=35\,\text{cm}^2\). 2. Thus, solids \(A\) and \(B\) have equal cross-sectional areas at every corresponding height and have the same height. 3. By Cavalieri's principle, their volumes are equal. Using a prism with base area \(35\,\text{cm}^2\) and height \(9\,\text{cm}\), \(V_A=V_B=35\cdot9=315\,\text{cm}^3\).

Both solids have volume \(315\,\text{cm}^3\).

- Determine how much of each cross section is actually material. - Subtract the shaft area from the outer area. - Compare the result with a prism having a constant cross-sectional area.

1. The material area in each horizontal cross section is \(64\,\text{m}^2-16\,\text{m}^2=48\,\text{m}^2\). 2. A solid prism with base area \(48\,\text{m}^2\) and height \(20\,\text{m}\) has the same cross-sectional area at every corresponding height. 3. By Cavalieri's principle, the material volume is \(V=48\cdot20=960\,\text{m}^3\).

The volume of material is \(960\,\text{m}^3\).

- Replace the offset duct with a right comparison prism. - Use the fact that corresponding cross sections have equal areas. - Convert the volume only after calculating it.

1. The offset duct has the same volume as a right prism with base area \(0.08\,\text{m}^2\) and height \(2.5\,\text{m}\). 2. The volume is \(V=0.08\cdot2.5=0.2\,\text{m}^3\). 3. Since \(1\,\text{m}^3=1{,}000\,\text{dm}^3\), \(0.2\,\text{m}^3=200\,\text{dm}^3\).

The interior volume is \(200\,\text{dm}^3\).

- What happens to the shape and area of a prism's cross section when the cutting plane stays parallel to the ends? - State both conditions needed for a Cavalieri comparison with a right prism. - Calculate the right prism's volume only after the comparison is justified.

1. Every cross section parallel to the ends of the oblique prism is congruent to an end, so it has area \(1.8\,\text{m}^2\). 2. A right prism with the same end area has the same cross-sectional area at every corresponding height, and both prisms have perpendicular height \(6\,\text{m}\). 3. Cavalieri's principle therefore gives equal volumes. 4. The right prism's volume is \(V=Bh=1.8\cdot6=10.8\,\text{m}^3\), so the oblique container has the same capacity.

Corresponding parallel cross sections have area \(1.8\,\text{m}^2\) and the prisms have the same perpendicular height, so Cavalieri's principle gives equal volumes. The capacity is \(10.8\,\text{m}^3\).

- Use the stated standard solid as a comparison. - The sculpture's outer shape does not matter when corresponding cross-sectional areas are equal. - Calculate the comparison solid's volume.

1. The sculpture and the cone have the same height and equal cross-sectional areas at every corresponding height, so Cavalieri's principle gives them equal volumes. 2. The cone's volume is \(V=\frac{1}{3}\pi\cdot6^2\cdot15=180\pi\,\text{cm}^3\approx565.49\,\text{cm}^3\).

The volume is \(180\pi\,\text{cm}^3\approx565.49\,\text{cm}^3\).

- Decide whether the redesign changes the area of each layer. - Compare the original and new layers at the same height. - Calculate the percent change only after comparing the volumes.

1. The original and redesigned components have equal heights and equal corresponding cross-sectional areas, so Cavalieri's principle gives equal volumes. 2. The new volume is \(2.4\,\text{m}^3\). 3. The volume change is \(2.4-2.4=0\,\text{m}^3\), so the percent change is \(0\%\).

The new volume is \(2.4\,\text{m}^3\), and the percent change is \(0\%\).

- Compare the constant cross-section scale factor with the volume scale factor. - Write an equation for the unknown value. - Check that the resulting factor is consistent with the larger volume.

1. Because the corresponding cross-sectional areas differ by a constant factor, the volumes differ by the same factor. 2. Therefore, \(\frac{x}{5}=1.4\). 3. Multiplying by \(5\) gives \(x=7\).

\(x=7\).

- Compare the two cross-sectional areas, not their outlines. - A right triangle has half the area of its corresponding base-height rectangle. - Every layer of each solid has the same area as its pictured section.

1. Solid \(K\) has section area \(8\cdot3=24\,\text{cm}^2\). 2. Solid \(L\) has section area \(\frac{1}{2}\cdot12\cdot4=24\,\text{cm}^2\). 3. Corresponding sections have equal areas over the same height, so Cavalieri's principle gives equal volumes. 4. Each volume is \(24\cdot15=360\,\text{cm}^3\).

\(V_K=V_L=360\,\text{cm}^3\).

- What can you infer about two constant cross-sectional areas from equal volumes and equal solid heights? - How do the labeled sides determine each triangle's area? - Your value of \(h\) should make the two cross-sectional areas agree.

1. Section a) has area \(\frac{1}{2}\cdot12\cdot5=30\,\text{cm}^2\). 2. The solids have equal volumes, equal heights, and constant sections, so those section areas must match. 3. Section b) has area \(\frac{1}{2}\cdot10h=5h\). Thus \(5h=30\) and \(h=6\,\text{cm}\).

\(h=6\,\text{cm}\). Equal volumes over the same height require the constant section areas to match.

- Replace the column with a simple solid having the same cross sections. - Work backward from volume to area.

1. The column has the same volume as a prism with unknown base area \(B\) and height \(18\,\text{cm}\). 2. From \(18B=936\), \(B=52\,\text{cm}^2\).

Each cross section has area \(52\,\text{cm}^2\).

- Compare a horizontal slice of each solid at the same height. - Determine whether the slice areas change as the slicing level moves upward. - Once Cavalieri gives equal volumes, use the known prism volume relationship.

1. Every horizontal cross-section of the cylinder is a circle of area \(\pi r^2\). 2. Every horizontal cross-section of the right prism is congruent to its base, so it also has area \(\pi r^2\). 3. The two solids have the same height and equal corresponding cross-sectional areas. Cavalieri's principle therefore gives them equal volumes. 4. The prism's volume is base area times height, \((\pi r^2)h\). Hence the cylinder's volume is \(V=\pi r^2h\).

Corresponding horizontal sections of the cylinder and prism both have area \(\pi r^2\), and the solids have the same height. By Cavalieri's principle, their volumes are equal. Since the prism has volume \((\pi r^2)h\), the cylinder has volume \(V=\pi r^2h\).

- Two slices may have different outlines. How could you compare the space each one covers? - Which information in each diagram matters for finding that amount of space? - What would stacking the slices through the given height tell you?

1. Section a) has area \(4\cdot3=12\,\text{cm}^2\). 2. Section b) has area \(6\cdot2=12\,\text{cm}^2\), although its shape is different. 3. Equal corresponding slice areas over the same height imply equal volumes. Both volumes are \(12\cdot7=84\,\text{cm}^3\).

Yes. Both volumes are \(84\,\text{cm}^3\).

- Compare the areas of the two end faces first. - Each cross section stays the same along its solid's length. - What does stacking equal-area slices through equal lengths give?

1. The rectangular section has area \(4\cdot3=12\,\text{cm}^2\). 2. The triangular section has area \(\frac{1}{2}\cdot6\cdot4=12\,\text{cm}^2\). 3. Both are extended through a perpendicular length of \(5\,\text{cm}\), so each volume is \(12\cdot5=60\,\text{cm}^3\).

Yes. Each has five layers of area \(12\,\text{cm}^2\) and thickness \(1\,\text{cm}\), so each volume is \(60\,\text{cm}^3\).

- Does moving a card sideways change how much space that card occupies? - Find the volume represented by just one layer. - What happens when all the layer volumes are added?

1. Each card occupies \(10\cdot0.5=5\,\text{cm}^3\). 2. All 20 cards together occupy \(20\cdot5=100\,\text{cm}^3\). 3. Shifting a layer horizontally does not change its area or thickness, so the volume stays \(100\,\text{cm}^3\).

No. Each stack has volume \(100\,\text{cm}^3\).

- Imagine a smaller similar cone or pyramid above the cut. - A length scale factor and an area scale factor are not the same. - The bases have equal areas, although one is round and one is polygonal.

1. In either solid, the cut is similar to the base with every length scaled by \(\frac{1}{2}\). 2. Areas scale by the square of the length factor, so each cut has area \(\bigl(\frac{1}{2}\bigr)^2\cdot20=5\,\text{cm}^2\). 3. The two solids therefore have matching section areas at that level, even though their section shapes differ.

Each cut has area \(5\,\text{cm}^2\), one-quarter of its base area.

- Sam has compared the bases. What might be different inside the two solids? - How does the cylinder's width change as you move upward? - How does the cone's width change over the same distance?

1. Halfway up, the cylinder has the same radius \(2\,\text{cm}\) as at its base, so its section area is \(4\pi\,\text{cm}^2\). 2. The cone's radius halfway up is \(1\,\text{cm}\), so its section area is \(\pi\,\text{cm}^2\). 3. The areas do not match, so the equal-slice condition for Cavalieri's principle fails. Equal base areas and total heights alone do not prove equal volumes. 4. In fact, the cylinder's volume is \(\pi\cdot2^2\cdot6=24\pi\,\text{cm}^3\), while the cone's is \(\frac{1}{3}\pi\cdot2^2\cdot6=8\pi\,\text{cm}^3\).

No. Halfway up, the cylinder's cross-sectional area is \(4\pi\,\text{cm}^2\), but the cone's is \(\pi\,\text{cm}^2\). The equal-slice condition fails. Their volumes are \(24\pi\,\text{cm}^3\) and \(8\pi\,\text{cm}^3\), respectively.

- Each smaller square has the same shape as the base. How does its distance from the tip compare with the full height? - Work out one square's side length before its area. - Compare the changes in the side lengths with the changes in the areas.

1. Each cut forms a square similar to the base. A cut at distance \(t\) from the tip has side length \(8(t/8)=t\,\text{cm}\). 2. At distances \(2\), \(4\), and \(6\,\text{cm}\), the square side lengths are \(2\), \(4\), and \(6\,\text{cm}\). 3. Their areas are \(4\), \(16\), and \(36\,\text{cm}^2\), respectively. 4. Equal increases in length do not give equal increases in area because square area depends on the square of the side length.

\(4\,\text{cm}^2\), \(16\,\text{cm}^2\), and \(36\,\text{cm}^2\). The increases are \(12\) and \(20\,\text{cm}^2\), not equal.

- First calculate \(B_1\) and \(B_2\) from the square edge lengths. - Consider what shape results when the top face shrinks to a single point. - Substitute zero for \(B_2\) in the general formula.

1. When \(b=4\,\text{cm}\), \(B_1=8^2=64\,\text{cm}^2\) and \(B_2=4^2=16\,\text{cm}^2\). Therefore, \(V=\frac{12}{3}\left(64+\sqrt{64\cdot16}+16\right)=4\left(64+32+16\right)=448\,\text{cm}^3\). 2. When \(b=0\,\text{cm}\), \(B_2=0\). Then \(V=\frac{12}{3}\left(64+\sqrt{64\cdot0}+0\right)=4\cdot64=256\,\text{cm}^3\). 3. When the top area shrinks to zero, the solid ends at an apex and becomes a pyramid. In general, setting \(B_2=0\) gives \(V=\frac{h}{3}(B_1+0+0)=\frac{1}{3}B_1h\).

1. \(V=448\,\text{cm}^3\) 2. \(V=256\,\text{cm}^3\) 3. The solid becomes a pyramid, and the formula becomes \(V=\frac{1}{3}B_1h\).

- Use the proportional relationship between the radii and heights of the similar cones. - The large cone's height includes both the frustum and the small cone. - Use the cone volume formula for each cone before subtracting.

1. Similar triangles give \(\frac{x}{4}=\frac{x+6}{8}\). 2. Solving, \(8x=4x+24\), so \(x=6\,\text{cm}\). The large cone's height is \(12\,\text{cm}\). 3. The cone volumes are \(V_{\text{large}}=\frac{1}{3}\pi\cdot8^2\cdot12=256\pi\,\text{cm}^3\) and \(V_{\text{small}}=\frac{1}{3}\pi\cdot4^2\cdot6=32\pi\,\text{cm}^3\). 4. Therefore, the frustum's volume is \(256\pi-32\pi=224\pi\,\text{cm}^3\approx703.72\,\text{cm}^3\). 5. Direct substitution gives \(V=\frac{6\pi}{3}\left(8^2+8\cdot4+4^2\right)=2\pi\left(64+32+16\right)=224\pi\,\text{cm}^3\), confirming the result.

a) \(x=6\,\text{cm}\) b) \(V=224\pi\,\text{cm}^3\approx703.72\,\text{cm}^3\) c) Direct substitution also gives \(224\pi\,\text{cm}^3\).

- What area do the base-parallel cross sections have? - Which height must match for Cavalieri's principle? - Use the diameter to find the radius before computing the comparison cylinder's volume. - Convert cubic centimeters to cubic decimeters at the end.

1. Every base-parallel cross section of the oblique cylinder is a circle congruent to its base, with area \(\pi(6)^2=36\pi\,\text{cm}^2\). The matching right-cylinder sections have the same area. 2. Both cylinders have perpendicular height \(15\,\text{cm}\), so Cavalieri's principle gives equal volumes. 3. The right-cylinder volume is \(V=\pi r^2h=\pi(6)^2(15)=540\pi\,\text{cm}^3\). 4. Since \(1000\,\text{cm}^3=1\,\text{dm}^3\), this is \(0.54\pi\,\text{dm}^3\approx1.70\,\text{dm}^3\). 5. The sloping side length does not change the perpendicular distance between the bases or the section areas.

Corresponding cross sections match, and the perpendicular heights match, so the volumes are equal. The oblique cylinder has volume \(0.54\pi\,\text{dm}^3\approx1.70\,\text{dm}^3\). The sloping side length is not needed.

- Compare cross sections taken the same distance from each apex. - Consider how cross-sectional area scales in a pyramid. - Once volume equality is established, calculate the volume only once.

1. At the same fractional distance from the apex, corresponding cross sections of all three pyramids have the same area scale factor. 2. Because the pyramids have equal base areas and equal heights, corresponding cross sections have equal areas. By Cavalieri's principle, the volumes are equal. 3. The common volume is \(V=\frac{1}{3}Bh=\frac{1}{3}\cdot48\cdot9=144\,\text{cm}^3\).

All three pyramids have volume \(144\,\text{cm}^3\).

- Think about how a parallel cross section of a cone is scaled relative to its base. - At the same fraction of the perpendicular height, compare the scale factors in the oblique and right cones. - Only after establishing equal corresponding areas should you use the right cone's volume formula.

1. In either cone, a cross section parallel to the base is similar to the base. Its radius is determined by the same linear scale factor from the apex because the two cones have the same base radius and perpendicular height. 2. Therefore, at every corresponding height, the two circular cross sections have equal radii and equal areas. 3. The cones have the same perpendicular height, so Cavalieri's principle gives them equal volumes. 4. The comparison right cone has volume \(V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(4.5)^2(12)=81\pi\,\text{cm}^3\approx254.47\,\text{cm}^3\).

Similarity makes the corresponding parallel cross sections equal in area at every height, so Cavalieri's principle gives equal volumes. The oblique cone's volume is \(81\pi\,\text{cm}^3\approx254.47\,\text{cm}^3\).

- Compare cross sections near the base and near the top for each pair. - Determine whether cross-sectional area changes with height. - Equal base areas alone are not sufficient.

1. For a), all corresponding cross sections have equal areas, so the volumes are equal. 2. For b), prism cross sections remain constant while pyramid cross sections decrease toward the apex, so Cavalieri's principle does not establish equal volume. 3. For c), corresponding cross sections of the right and oblique cones have equal areas, so the volumes are equal. 4. For d), cylinder and cone cross sections do not have equal areas at every height, so the principle does not establish equal volume.

a) yes b) no c) yes d) no

- Simplify and compare the area expressions pairwise. - Look for a cross-sectional area difference that remains constant. - Interpret a constant area difference as the cross section of an added prism.

1. For every height, \(A_K(z)=A_L(z)\), so Cavalieri's principle gives \(V_K=V_L\). 2. Also, \(A_M(z)-A_K(z)=1\,\text{cm}^2\) and \(A_M(z)-A_L(z)=1\,\text{cm}^2\) at every height. 3. This constant cross-sectional area difference over \(10\,\text{cm}\) corresponds to a volume difference of \(1\cdot10=10\,\text{cm}^3\). 4. Therefore, \(V_M\) is \(10\,\text{cm}^3\) greater than each of \(V_K\) and \(V_L\).

\(V_K=V_L\). The volume \(V_M\) is \(10\,\text{cm}^3\) greater than each of them.

- Subtract the inner circular area from the outer circular area for one parallel cross section. - Ask whether that annular area changes as the cross section moves through the oblique pipe. - Match both cross-sectional area and perpendicular height to a right hollow cylinder before calculating volume.

1. Every cross section parallel to the ends is an annulus with area \(A=\pi(5^2-3^2)=16\pi\,\text{cm}^2\). 2. A right hollow cylinder with outer radius \(5\,\text{cm}\), inner radius \(3\,\text{cm}\), and height \(18\,\text{cm}\) has the same annular area in every corresponding cross section. 3. The two solids have the same perpendicular height, so Cavalieri's principle gives equal volumes. 4. Thus, \(V=(16\pi)(18)=288\pi\,\text{cm}^3\approx904.78\,\text{cm}^3\).

Each parallel cross section has area \(16\pi\,\text{cm}^2\). These areas match those of a right hollow cylinder at every corresponding height, so Cavalieri's principle applies. The material volume is \(288\pi\,\text{cm}^3\approx904.78\,\text{cm}^3\).

- Identify a familiar solid with the same cross-sectional area at every height. - The changing cross-sectional shape does not matter if the area remains equal. - Calculate in one unit first, then convert at the end.

1. The sculpture has the same volume as a cylinder with radius \(6\,\text{cm}\) and height \(20\,\text{cm}\). 2. Its volume is \(V=\pi\cdot6^2\cdot20=720\pi\,\text{cm}^3\). 3. Since \(1{,}000\,\text{cm}^3=1\,\text{dm}^3\), \(720\pi\,\text{cm}^3=0.72\pi\,\text{dm}^3\approx2.26\,\text{dm}^3\).

The volume is \(0.72\pi\,\text{dm}^3\approx2.26\,\text{dm}^3\).

- Compare cross sections at the same perpendicular distance from the base. - Decide whether shifting the apex changes the area scale factor of those cross sections. - Calculate the common volume only once.

1. At the same perpendicular distance from the base, a cross section parallel to the base has the same area in both pyramids. Shifting the apex sideways changes the position of the cross section, but not its area scale factor. 2. The pyramids have equal heights and equal corresponding cross-sectional areas, so Cavalieri's principle shows that their volumes are equal. 3. Their common volume is \(V=\frac{1}{3}\cdot80\cdot9=240\,\text{cm}^3\).

Both pyramids have volume \(240\,\text{cm}^3\).

- What shape and area does each slice parallel to the bases have? - Does leaning the cylinder change the area of these slices or the perpendicular height? - Once you have the correct volume, compare it with Tim's expression.

1. Every slice parallel to the bases is a circle of area \(25\pi\,\text{cm}^2\) in either cylinder. Their perpendicular heights match, so Cavalieri's principle gives equal volumes. 2. The correct volume is \(V=\pi(5)^2(12)=300\pi\,\text{cm}^3\approx942.48\,\text{cm}^3\). 3. Tim used the sloping length instead of the perpendicular height. His \(325\pi\,\text{cm}^3\) is too large by \(25\pi\,\text{cm}^3\approx78.54\,\text{cm}^3\).

Equal circular slices and equal perpendicular heights give the same volume as a right cylinder. Correct volume: \(300\pi\,\text{cm}^3\). Overestimate: \(25\pi\,\text{cm}^3\approx78.54\,\text{cm}^3\).

- Compare cross sections near the base and near the top. - Determine whether the cross-sectional area stays constant in both solids. - Equal base area and equal height alone are not sufficient.

1. The corresponding cross-sectional areas are not equal at every height. A prism's cross-sectional area remains constant, while a pyramid's cross-sectional area decreases toward the apex. 2. Therefore, equal base area and equal height do not satisfy the cross-section condition of Cavalieri's principle for these two solids. 3. The prism's volume is \(V_{\text{prism}}=54\cdot10=540\,\text{cm}^3\). 4. The pyramid's volume is \(V_{\text{pyramid}}=\frac{1}{3}\cdot54\cdot10=180\,\text{cm}^3\).

Nora's reasoning is incorrect. The prism has volume \(540\,\text{cm}^3\), and the pyramid has volume \(180\,\text{cm}^3\).

- Treat the difference between the two cross sections as a separate simple solid. - This area difference remains constant over the entire height. - Add the resulting extra volume to the known volume.

1. The constant cross-sectional area difference can be represented by an added prism with base area \(7\,\text{cm}^2\) and height \(12\,\text{cm}\). 2. The added volume is \(7\cdot12=84\,\text{cm}^3\). 3. Therefore, \(V_A=420+84=504\,\text{cm}^3\).

The volume of solid \(A\) is \(504\,\text{cm}^3\).

- The dashed line gives the perpendicular height of the trapezoid. - Compare cross-sectional areas first. - Matching areas at every level and matching solid heights imply matching volumes.

1. Trapezoid a) has section area \(\frac{10+6}{2}\cdot4=32\,\text{cm}^2\). 2. Rectangle b) has section area \(8\cdot4=32\,\text{cm}^2\). 3. The equal areas continue through the same height, so both volumes are \(32\cdot9=288\,\text{cm}^3\).

\(V_a=V_b=288\,\text{cm}^3\).

- The white center is empty space, not material. - A framed cross section can have the same area as a solid square. - Compare material areas, then multiply by the common height.

1. Section a) has material area \(12\cdot8-8\cdot4=96-32=64\,\text{cm}^2\). 2. Section b) has area \(8\cdot8=64\,\text{cm}^2\). 3. Corresponding material sections match in area across the common height, so each volume is \(64\cdot20=1280\,\text{cm}^3\).

Both contain \(1280\,\text{cm}^3\) of material.

- Compute each section area with its correct shape formula. - Only the right triangle uses the one-half factor. - Compare areas at corresponding levels before concluding anything about volumes.

1. Their cross-sectional areas are \(A_a=5\cdot5=25\,\text{cm}^2\), \(A_b=10\cdot2.5=25\,\text{cm}^2\), and \(A_c=\frac{1}{2}\cdot12\cdot5=30\,\text{cm}^2\). 2. A and B have matching section areas at every height and the same height, so their volumes are equal. 3. \(V_a=V_b=25\cdot12=300\,\text{cm}^3\); \(V_c=30\cdot12=360\,\text{cm}^3\).

a) \(300\,\text{cm}^3\). b) \(300\,\text{cm}^3\). c) \(360\,\text{cm}^3\). Solids a) and b) have equal volumes.

- Distinguish the outer region from the hollow space. - Compare the resulting areas symbolically. - Keep the factor \(\pi\) until the final approximation.

1. The annular material area in a) is \(\pi\left(5^2-3^2\right)=16\pi\,\text{cm}^2\). 2. The circular material area in b) is \(\pi\cdot4^2=16\pi\,\text{cm}^2\). 3. The material cross sections have equal areas at every height, so Cavalieri's principle gives \(V=(16\pi)(11)=176\pi\,\text{cm}^3\approx552.92\,\text{cm}^3\).

Both material volumes are \(176\pi\,\text{cm}^3\approx552.92\,\text{cm}^3\).

- Does each picture represent one connected region, or more than one? - Which two lengths can you use to find the area of a right triangle? - What would equal total slice areas at every height tell you?

1. Solid \(A\) has section area \(\frac{1}{2}\cdot12\cdot8=48\,\text{cm}^2\). 2. Each separate triangle in b) has area \(\frac{1}{2}\cdot6\cdot8=24\,\text{cm}^2\). The two together have area \(48\,\text{cm}^2\). 3. Equal total areas in corresponding slices and equal solid heights imply equal volumes. 4. Each volume is \(48\cdot10=480\,\text{cm}^3\).

\(V_A=V_B=480\,\text{cm}^3\).

- You know the total volumes and the solid heights are equal. - Constant-section solids have volume equal to section area times height. - Use the triangle to recover the area of the rectangle.

1. Since each solid has constant section area over the same height, equal volumes imply equal section areas. 2. The right triangle has area \(\frac{1}{2}\cdot15\cdot4=30\,\text{cm}^2\). 3. The rectangle has area \(6x\), so \(6x=30\) and \(x=5\,\text{cm}\). 4. Their common volume is \(30\cdot13=390\,\text{cm}^3\).

\(x=5\,\text{cm}\). Both volumes are \(390\,\text{cm}^3\).

- Only the colored part of a) counts as material. - Compare material areas, since both columns have the same height and constant sections. - A side length must be positive after solving for its square.

1. Equal volumes and heights, with constant sections, imply equal material section areas. 2. Section a) has material area \(10^2-x^2\), and b) has area \(8^2\). 3. Equating gives \(100-x^2=64\), so \(x^2=36\) and \(x=6\,\text{cm}\), using the positive length. 4. Each material volume is \(64\cdot15=960\,\text{cm}^3\).

\(x=6\,\text{cm}\). Each contains \(960\,\text{cm}^3\) of material.

- Use the area formula that matches each cross-section shape. - What common factor appears in the two volumes? - Think about whether moving a thin horizontal layer sideways changes its volume.

1. Rectangle \(A\) has area \(9\cdot4=36\,\text{cm}^2\); right triangle \(B\) has area \(\frac{1}{2}\cdot12\cdot9=54\,\text{cm}^2\). 2. Since the heights match and sections are constant, the volume ratio is \(36:54=2:3\). 3. Their volumes are \(V_A=36\cdot8=288\,\text{cm}^3\) and \(V_B=54\cdot8=432\,\text{cm}^3\). 4. Shifting layers horizontally changes neither layer area nor the number and thickness of layers, so the volumes and their ratio stay the same.

\(V_A:V_B=2:3\). Shifting the layers sideways changes neither their areas nor their thicknesses, so the ratio stays the same.

- The half in the triangle area formula does not compare two triangles with different dimensions automatically. - Use the dashed perpendicular height in a). - Check the three section areas before comparing total volumes.

1. The parallelogram area is \(8\cdot5=40\,\text{cm}^2\). The rectangle area is \(10\cdot4=40\,\text{cm}^2\). 2. The right triangle's area is \(\frac{1}{2}\cdot10\cdot8=40\,\text{cm}^2\). 3. Ben's claim is false: the triangle has a larger perpendicular height, offsetting the one-half factor. 4. Corresponding section areas agree through equal solid heights, so all three volumes are \(40\cdot7=280\,\text{cm}^3\).

Ben is incorrect. All three volumes are \(280\,\text{cm}^3\).

- Use a separate area formula for each distinct shape. - A sloping side is not necessarily the perpendicular height of a trapezoid or parallelogram. - Any candidate with a different cross-sectional area has a different volume at the same solid height.

1. Reference section area is \(7\cdot4=28\,\text{cm}^2\). 2. Candidate a) has area \(\frac{1}{2}\cdot14\cdot4=28\,\text{cm}^2\); b) has area \(\frac{9+5}{2}\cdot4=28\,\text{cm}^2\); c) has area \(10\cdot2.5=25\,\text{cm}^2\). 3. Reference \(R\) and candidates a), b) have equal section areas at all heights, so their volumes are \(28\cdot9=252\,\text{cm}^3\). 4. Candidate c) has volume \(25\cdot9=225\,\text{cm}^3\).

Candidates a) and b) match \(R\). \(V_R=V_a=V_b=252\,\text{cm}^3\); \(V_c=225\,\text{cm}^3\).

- Start with the areas of the two pictured sections. - Imagine building the known solid from thin layers. How would you build the other solid with the same total volume? - What equation connects the unknown height with its slice area?

1. Section A has area \(12\cdot3=36\,\text{cm}^2\), while section B has area \(\frac{8+4}{2}\cdot5=30\,\text{cm}^2\). 2. Each solid matches a prism with the same constant section area and height, so \(V_A=36\cdot10=360\,\text{cm}^3\) and \(V_B=30h\). 3. Equal volumes require \(30h=360\), giving \(h=12\,\text{cm}\).

Solid B must be \(12\,\text{cm}\) tall.

- Compare cross-sections taken at the same relative height in the two solids. - When all linear dimensions scale by a factor, determine how area scales. - After establishing equal corresponding slice areas, connect the cone to the known pyramid volume formula.

1. At a fraction \(t\) of the distance from the apex to the base, linear dimensions in each cross-section are scaled by \(t\) from the corresponding base dimensions. 2. Therefore, each cross-sectional area is scaled by \(t^2\). Because the cone and pyramid have the same base area \(B\), their corresponding cross-sections both have area \(t^2B\). 3. The solids have the same height and equal corresponding cross-sectional areas, so Cavalieri's principle gives them equal volumes. 4. The pyramid has volume \(\frac{1}{3}Bh\). Therefore the cone also has volume \(\frac{1}{3}Bh\). 5. Since \(B=\pi r^2\), the cone formula is \(V=\frac{1}{3}\pi r^2h\).

Similarity makes every pair of corresponding cross-sections have the same area \(t^2B\). By Cavalieri's principle, the cone and matching pyramid have equal volume, so \(V=\frac{1}{3}Bh=\frac{1}{3}\pi r^2h\).

- Find the cube's volume, then use the stated six equal pieces. - The pyramid's perpendicular height is not a full cube edge. Where is its tip? - The matching prism has the same base and height as the pyramid. Does it fill the entire cube?

1. The cube has volume \(6^3=216\,\text{cm}^3\). Six equal pieces each have volume \(216\div6=36\,\text{cm}^3\). 2. A pyramid's base is a cube face, so \(B=6^2=36\,\text{cm}^2\). Its height is the center-to-face distance, \(h=3\,\text{cm}\). 3. The formula gives \(\frac{1}{3}Bh=\frac{1}{3}\cdot36\cdot3=36\,\text{cm}^3\). 4. A prism with the same base and height occupies \(36\cdot3=108\,\text{cm}^3\), only half the whole cube. Thus one pyramid is one-third of that prism but one-sixth of the cube.

a) \(36\,\text{cm}^3\) b) \(B=36\,\text{cm}^2\), \(h=3\,\text{cm}\), so \(\frac{1}{3}Bh=36\,\text{cm}^3\). c) The matching prism has half the cube's volume, so one-sixth of the cube equals one-third of that prism.

- The upper piece is similar to the original pyramid. Which lengths have changed? - A length scale factor does not also give the area or volume scale factor. - The two parts together must account for the whole pyramid.

1. The upper pyramid is similar to the whole pyramid, with every length scaled by \(\frac{1}{2}\). 2. Its cross-sectional lengths are half as long, its base area is one-quarter, and its height is one-half; its volume is \((\frac{1}{2})^3=\frac{1}{8}\) of the whole. 3. The lower part has the remaining fraction \(1-\frac{1}{8}=\frac{7}{8}\). 4. Lea confused halving height with halving volume. A pyramid's cross-sectional area changes as it narrows toward the tip.

No. The upper pyramid has \(\frac{1}{8}\) of the volume; the lower part has \(\frac{7}{8}\).

- Use a right triangle to find the sphere's section radius at the given height. - Use similarity to find the radius of the cone's section at the same height. - How much evidence about the other heights has Noah provided?

1. The sphere's section radius is \(\sqrt{5^2-3^2}=4\,\text{cm}\), so the area is \(16\pi\,\text{cm}^2\). 2. Similar triangles show that the removed cone has section radius \(3\,\text{cm}\) at this height. The remaining cylinder area is \(25\pi-9\pi=16\pi\,\text{cm}^2\). 3. Noah checked only one height. Cavalieri's principle requires equal corresponding section areas at every height, not just one.

Both sections have area \(16\pi\,\text{cm}^2\) at that height. One matching level is not enough to conclude that volumes match.

- Match each square base area with the corresponding circular base area. - Think about how a side length or radius changes linearly between the two ends of a frustum. - Compare cross-sectional areas at the same relative height before transferring a volume formula from one solid to the other.

1. Equal lower-base areas give \(a_1^2=\pi r_1^2\), so \(r_1=\frac{a_1}{\sqrt{\pi}}\). Equal upper-base areas give \(a_2^2=\pi r_2^2\), so \(r_2=\frac{a_2}{\sqrt{\pi}}\). 2. At fraction \(t\) of the height from the lower base, the square cross-section has side length \((1-t)a_1+ta_2\), while the circular cross-section has radius \((1-t)r_1+tr_2\). 3. Since \(a_1=\sqrt{\pi}r_1\) and \(a_2=\sqrt{\pi}r_2\), the square cross-sectional area is \(\left((1-t)a_1+ta_2\right)^2=\pi\left((1-t)r_1+tr_2\right)^2\), exactly the area of the circular cross-section. 4. The solids have the same height and equal cross-sectional areas at every height, so Cavalieri's principle shows that their volumes are equal. 5. For the square pyramidal frustum, \(B_1=a_1^2=\pi r_1^2\) and \(B_2=a_2^2=\pi r_2^2\). Substituting gives \(V=\frac{h}{3}\left(\pi r_1^2+\sqrt{\pi^2r_1^2r_2^2}+\pi r_2^2\right)\). 6. Because the radii are nonnegative, \(\sqrt{\pi^2r_1^2r_2^2}=\pi r_1r_2\). Therefore, \(V=\frac{\pi h}{3}\left(r_1^2+r_1r_2+r_2^2\right)\).

a) \(r_1=\frac{a_1}{\sqrt{\pi}}\) and \(r_2=\frac{a_2}{\sqrt{\pi}}\) b) Corresponding parallel cross-sections have equal area, so Cavalieri's principle gives equal volumes. Hence \(V=\frac{\pi h}{3}\left(r_1^2+r_1r_2+r_2^2\right)\).

- Rewrite both volume formulas in terms of \(r_1\), \(r_2\), and \(h\). - Compare the two expressions by studying their difference. - Determine when that difference can equal zero.

1. Substituting \(r_1=r_2=r\) gives \(V=\frac{1}{3}\pi h(r^2+r^2+r^2)=\pi r^2h\), the cylinder volume formula. 2. The proposed formula becomes \(V_{\text{estimate}}=\pi h\left(\frac{r_1+r_2}{2}\right)^2=\frac{\pi h}{4}(r_1^2+2r_1r_2+r_2^2)\). 3. Subtracting the estimate from the exact formula gives \(V-V_{\text{estimate}}=\frac{\pi h}{12}\left(4r_1^2+4r_1r_2+4r_2^2-3r_1^2-6r_1r_2-3r_2^2\right)\). 4. Therefore, \(V-V_{\text{estimate}}=\frac{\pi h}{12}(r_1-r_2)^2\). This difference is positive when \(r_1\ne r_2\), so the proposed cylinder formula is not exact and underestimates the frustum volume.

1. The formula simplifies to \(V=\pi r^2h\). 2. The proposed formula is not exact when \(r_1\ne r_2\). It underestimates the volume by \(\frac{\pi h}{12}(r_1-r_2)^2\).

- Identify the segment in the cube that becomes a tetrahedron edge. - Use the general volume formula for a pyramid. - Determine the shape and area of a corner pyramid's base. - Solve the edge-length relationship for \(a\) before substituting.

1. Each tetrahedron edge is a face diagonal of the cube. By the Pythagorean theorem, \(b^2=a^2+a^2=2a^2\), so \(b=a\sqrt{2}\). 2. A corner pyramid has a right-triangular base with area \(\frac{1}{2}a^2\) and perpendicular height \(a\). Thus, \(V_{\text{corner}}=\frac{1}{3}\cdot\frac{1}{2}a^2\cdot a=\frac{1}{6}a^3\). 3. Subtracting the four corner pyramids from the cube gives \(V_{\text{tetrahedron}}=a^3-4\cdot\frac{1}{6}a^3=\frac{1}{3}a^3\). 4. From \(b=a\sqrt{2}\), \(a=\frac{b}{\sqrt{2}}\). Therefore, \(V=\frac{1}{3}\left(\frac{b}{\sqrt{2}}\right)^3=\frac{b^3}{6\sqrt{2}}=\frac{\sqrt{2}}{12}b^3\).

a) \(b=a\sqrt{2}\) b) \(V_{\text{corner}}=\frac{1}{6}a^3\) and \(V_{\text{tetrahedron}}=\frac{1}{3}a^3\) c) \(V=\frac{\sqrt{2}}{12}b^3\)

- Consider how a height changes when a solid is turned upside down. - Convert complementary heights into corresponding heights. - Changing a solid's orientation does not change its volume.

1. Turn solid \(L\) upside down. A cross section that was originally at height \(12-z\) is then at height \(z\). 2. Therefore, at every corresponding height \(z\), solid \(K\) and the inverted solid \(L\) have equal cross-sectional areas. 3. Turning a solid over does not change its volume, so Cavalieri's principle gives \(V_K=V_L\).

After solid \(L\) is turned upside down, all corresponding cross-sectional areas are equal. Therefore, \(V_K=V_L\).

- Add the two cross-sectional areas at the same height. - Check whether the height-dependent terms cancel. - Interpret the resulting total cross-sectional area using a comparison solid.

1. At each height, the sum of the cross-sectional areas is \(A_A(z)+A_B(z)=10+2z+28-2z=38\,\text{cm}^2\). 2. Together, the two solids have the same total cross-sectional area at every height as a prism with base area \(38\,\text{cm}^2\) and height \(9\,\text{cm}\). 3. Therefore, \(V_A+V_B=38\cdot9=342\,\text{cm}^3\).

\(V_A+V_B=342\,\text{cm}^3\).

- Calculate the two volumes independently. - Then compare at least one pair of corresponding cross sections. - A true implication does not necessarily remain true when reversed.

1. The prism's volume is \(V_{\text{prism}}=30\cdot10=300\,\text{cm}^3\). 2. The pyramid's volume is \(V_{\text{pyramid}}=\frac{1}{3}\cdot90\cdot10=300\,\text{cm}^3\). 3. However, their base cross-sectional areas are already different: \(30\,\text{cm}^2\) for the prism and \(90\,\text{cm}^2\) for the pyramid. 4. Thus, equal height and equal volume do not imply equal cross-sectional areas at every corresponding height. The converse is false.

The claim is false. Both volumes are \(300\,\text{cm}^3\), but the corresponding cross-sectional areas are not equal at every height.

- Follow a line from \(A\) through an interior point until it reaches a face. Which pyramid contains that line? - Can you rotate the cube around its long diagonal to exchange the three faces away from \(A\)? - Compare a cube's full edge length with the distance from a corner to its opposite face.

1. Each point inside the cube lies in the pyramid leading from \(A\) to the farthest of the three face directions from \(A\). Boundaries can belong to more than one piece, but their interiors do not overlap. Together, the pieces fill the cube. 2. Rotating the cube to exchange the three face directions also exchanges the pyramids. The three pieces have equal volumes. 3. The cube has volume \(6^3=216\,\text{cm}^3\), so each pyramid has volume \(216\div3=72\,\text{cm}^3\). 4. For cube side \(s\), each pyramid has base area \(B=s^2\) and perpendicular height \(h=s\). One-third of the cube is \(s^3/3=\frac{1}{3}Bh\).

a) The three congruent pyramids fill the cube, so each has one-third of its volume. b) \(72\,\text{cm}^3\) c) \(B=s^2\), \(h=s\), and \(V=\frac{s^3}{3}=\frac{1}{3}Bh\).

- Find a triangular face shared by two adjacent pieces. - In a parallelogram, what can you say about the point where its diagonals cross? - How could that point help you compare the distances of opposite vertices from a plane? - After establishing the volume of a triangular pyramid, think about building a polygon from triangles.

1. The three pyramids form a dissection of the prism: together they fill it without overlaps. 2. In parallelogram \(BCFE\), diagonals \(BF\) and \(CE\) bisect each other. The midpoint of \(CE\) lies on \(BF\) and hence in plane \(ABF\), so \(C\) and \(E\) lie at equal perpendicular distances on opposite sides of that plane. Pyramids \(ABCF\) and \(ABEF\) share base \(ABF\) and have equal heights. Their sections at matching distances from that base have the same scale and area, so Cavalieri's principle gives equal volumes. 3. In parallelogram \(ABED\), diagonals \(AE\) and \(BD\) bisect each other. The midpoint of \(BD\) lies in plane \(AEF\), so \(B\) and \(D\) are equally far from it. Pyramids \(ABEF\) and \(ADEF\) share base \(AEF\) and have equal heights; the same equal-slice argument gives equal volumes. 4. The prism has volume \(Bh\). Because its three pyramid pieces have equal volume, each is \(\frac{1}{3}Bh\). In particular, \(ADEF\) has triangular base \(DEF\) of area \(B\) and height \(h\), so this gives the formula for a triangular pyramid. 5. Any polygonal base can be split into triangles. The pyramids from those triangles share the same apex and height, so their volumes add to \(\frac{1}{3}(B_1+B_2+\cdots)h=\frac{1}{3}Bh\).

a) In parallelogram \(BCFE\), diagonals \(BF\) and \(CE\) bisect each other. Thus \(C\) and \(E\) are equally far from shared base plane \(ABF\). In parallelogram \(ABED\), diagonals \(AE\) and \(BD\) similarly put \(B\) and \(D\) equally far from shared base plane \(AEF\). Each adjacent pair has the same triangular base and perpendicular height, so its parallel slices have equal areas. Cavalieri's principle makes all three volumes equal. b) The three equal-volume pieces fill the prism of volume \(Bh\), so each has volume \(\frac{1}{3}Bh\). A polygonal base can be split into triangles; adding the volumes of pyramids with those triangular bases and common height gives \(\frac{1}{3}Bh\) for any polygonal pyramid.

- In the sphere's axial circle, relate the slice radius, the distance \(y\), and the sphere radius with a right triangle. - In the comparison solid, subtract the cone's circular slice from the cylinder's circular slice. - Establish equality of corresponding slice areas before comparing total volumes. - The two removed cones are congruent and together occupy both halves of the cylinder.

1. In the sphere, the radius \(x\) of the circular slice satisfies \(x^2+y^2=r^2\), so its area is \(\pi x^2=\pi(r^2-y^2)\). 2. At the same height in the comparison solid, the cylinder slice has area \(\pi r^2\). The removed cone at that level has slice radius \(|y|\), so its slice area is \(\pi y^2\). 3. The remaining cross-sectional area is therefore \(\pi r^2-\pi y^2=\pi(r^2-y^2)\), exactly matching the sphere slice. 4. The solids have equal height and equal corresponding cross-sectional areas, so Cavalieri's principle gives them equal volumes. 5. The cylinder has volume \(\pi r^2(2r)=2\pi r^3\). The two removed cones have total volume \(2\left(\frac{1}{3}\pi r^2(r)\right)=\frac{2}{3}\pi r^3\). 6. Hence the sphere volume is \(2\pi r^3-\frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3\).

At height \(y\), both solids have cross-sectional area \(\pi(r^2-y^2)\). Cavalieri's principle therefore makes their volumes equal, giving \(V=2\pi r^3-\frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3\).