Volumes with cross-sections

Learning goals

Identifies the cross-sectional area as a function of position.
Sets up V = ∫A(x)dx or the corresponding y-form.
Uses correct limits and geometric formulas for the cross-sections.
Evaluates the integral and interprets the result as volume.

Difficulty: 1 – 5

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Volumes with cross-sections 1

A solid extends from \(x=0\) to \(x=3\), where \(x\) is measured in inches. Its cross-sectional area perpendicular to the x-axis is \(A(x)=5+2x-x^2\) square inches. Find the volume.

Hints

- The area function is already provided, so no base-width conversion is needed. - Integrating square units over inches gives cubic inches.

Solution

1. Volume accumulates cross-sectional area: \(V=\int_0^3A(x)\,\mathrm{d}x\). 2. \(V=\int_0^3(5+2x-x^2)\,\mathrm{d}x=15\,\text{in}^3\).

Answer

\(15\,\text{in}^3\)

Volumes with cross-sections 2

A solid extends from \(x=0\) to \(x=2\). Each perpendicular cross section is a circle. The radius of a representative cross section is labeled in the diagram. Find the volume.

Hints

- Read the radius expression from the representative cross section. - Use that radius in the circle-area formula. - Accumulate the circular areas from \(x=0\) to \(x=2\).

Solution

1. The diagram gives the radius \(r(x)=2-x\). 2. The cross-sectional area is \(A(x)=\pi(2-x)^2\). 3. Therefore, \(V=\pi\int_0^2(2-x)^2\,\mathrm{d}x=\frac{8\pi}{3}\).

Answer

\(\frac{8\pi}{3}\) cubic units

Volumes with cross-sections 3

The graph shows a base region and a representative vertical segment at \(x=0.5\). Cross sections perpendicular to the x-axis are squares. Find the side length and area of the square cross section at \(x=0.5\).

Hints

- Read the vertical distance across the base at the marked x-value. - That distance is a side length, not yet an area. - How do you convert a square's side length into its area?

Solution

1. At \(x=0.5\), the displayed vertical segment runs from \(y=0\) to \(y=1.5\), so its length is \(\frac{3}{2}\) units. 2. That segment is the side of the square cross section, so the area is \(\left(\frac{3}{2}\right)^2=\frac{9}{4}\) square units.

Answer

Side length: \(\frac{3}{2}\) units Cross-sectional area: \(\frac{9}{4}\) square units

Volumes with cross-sections 4

A semicircular cross section is one half of the displayed circle, and the labeled segment is its diameter. Find the area of the semicircular cross section.

Hints

- The labeled segment gives the full diameter, not the radius. - How are diameter and radius related? - A semicircle uses half of the corresponding circle's area.

Solution

1. The displayed diameter is \(6\,\text{cm}\), so the radius is \(3\,\text{cm}\). 2. A semicircle has half the area of a circle, so \(A=\frac{1}{2}\pi(3)^2=\frac{9\pi}{2}\,\text{cm}^2\).

Answer

\(\frac{9\pi}{2}\,\text{cm}^2\)

Volumes with cross-sections 5

A solid has cross sections perpendicular to the x-axis. Near \(x=2\,\text{cm}\), each cross section has area about \(6\,\text{cm}^2\). Estimate the volume of a thin slab from \(x=2\,\text{cm}\) to \(x=2.5\,\text{cm}\).

Hints

- A thin slab's volume is approximately its cross-sectional area times its thickness. - Find the thickness from the two given x-coordinates.

Solution

1. The slab has thickness \(2.5-2=0.5\,\text{cm}\). 2. Approximating it as a prism gives \(V\approx6\,\text{cm}^2\cdot0.5\,\text{cm}=3\,\text{cm}^3\).

Answer

About \(3\,\text{cm}^3\).

Volumes with cross-sections 6

Imagine slicing the box perpendicular to its longest edge. What shape is one cross section, and what is its area?

Hints

- Picture a cut across the box rather than along its longest side. - Use the two edge lengths perpendicular to the direction of the cut.

Solution

1. The longest edge is \(4\,\text{cm}\). A perpendicular slice cuts a rectangle with sides \(2\,\text{cm}\) and \(3\,\text{cm}\). 2. Its area is \(2\cdot3=6\,\text{cm}^2\).

Answer

A rectangle with area \(6\,\text{cm}^2\).

Volumes with cross-sections 7

The shaded base is the triangular region bounded by \(y=0\), \(y=2-x\), and \(x=0\). Cross sections perpendicular to the x-axis are regular hexagons whose short diagonals equal the vertical base width. Use \(A=\frac{\sqrt3}{2}d^2\) for a regular hexagon with short diagonal \(d\). Find the volume.

Hints

- Use the vertical width of the triangular base as the short diagonal. - Substitute that variable diagonal into the supplied hexagon-area formula. - Integrate from \(x=0\) to the x-intercept of \(y=2-x\).

Solution

1. The short diagonal is the vertical base width \(d(x)=2-x\). 2. The cross-sectional area is \(A(x)=\frac{\sqrt3}{2}(2-x)^2\). 3. Therefore, \(V=\frac{\sqrt3}{2}\int_0^2(2-x)^2\,\mathrm{d}x=\frac{4\sqrt3}{3}\).

Answer

\(\frac{4\sqrt3}{3}\) cubic units

Volumes with cross-sections 8

Two solids have the same base region and the same slicing interval. Solid S has square cross sections whose sides equal the base width. Solid C has circular cross sections whose diameters equal that same width. Compare their volumes without knowing the base curves.

Hints

- Compare the two cross-sectional areas for the same width. - The ratio between the areas is constant at every slice. - A constant cross-sectional area ratio gives the same volume ratio.

Solution

1. At each slice, \(A_S=w^2\). 2. At the same slice, \(A_C=\frac{\pi}{4}w^2\). 3. Therefore, \(A_C=\frac{\pi}{4}A_S\) at every position. 4. Integrating gives \(V_C=\frac{\pi}{4}V_S\). Since \(\frac{\pi}{4}<1\), Solid S has the larger volume.

Answer

\(V_C=\frac{\pi}{4}V_S\); Solid S has the larger volume.

Volumes with cross-sections 9

A base has vertical width \(w(x)=2\sin x\) on \([0,\pi]\), and perpendicular cross sections are squares. Andrei writes \(V=\int_0^\pi4w(x)\,\mathrm{d}x\). Identify the error and find the correct volume.

Hints

- What two-dimensional quantity must be accumulated to obtain volume? - How is the area of a square related to its side length \(w(x)\)? - After correcting the area function, integrate it over the given slicing interval.

Solution

1. The expression \(4w(x)\) is the square's perimeter, not its area. 2. The correct cross-sectional area is \(A(x)=w(x)^2=4\sin^2x\). 3. Therefore, \(V=\int_0^\pi4\sin^2x\,\mathrm{d}x=2\pi\).

Answer

Andrei integrated perimeter. The correct volume is \(2\pi\) cubic units.

Volumes with cross-sections 10

For \(0\le x\le9\), each cross section of a solid is a triangle with base \(\sqrt{x}\) and constant altitude \(5\). Find the volume.

Hints

- Use the two distinct triangle dimensions exactly as given. - Only the base varies with position.

Solution

1. The cross-sectional area is \(A(x)=\frac12(\sqrt x)(5)=\frac52\sqrt x\). 2. \(V=\int_0^9\frac52\sqrt x\,\mathrm{d}x=45\).

Answer

\(45\) cubic units

Volumes with cross-sections 11

A base has vertical width \(w(x)=\cos x\) for \(-\frac{\pi}{2}\le x\le\frac{\pi}{2}\). Perpendicular cross sections are rectangles whose height is twice the base width. Find the volume.

Hints

- Use \(w(x)\) and \(2w(x)\) as the two rectangle dimensions. - Multiply the dimensions before integrating. - Use symmetry or a power-reduction identity to evaluate the cosine-squared integral.

Solution

1. Each cross-sectional area is \(A(x)=w(x)\cdot2w(x)=2\cos^2x\). 2. Therefore, \(V=\int_{-\pi/2}^{\pi/2}2\cos^2x\,\mathrm{d}x=\pi\).

Answer

\(\pi\) cubic units

Volumes with cross-sections 12

A base has vertical width \(w(x)=e^{-x^2}\) on \([0,2]\). Cross sections perpendicular to the x-axis are squares. Use a calculator to find the volume to three decimal places.

Hints

- Square the entire width function. - Set up the exact integral before using numerical technology.

Solution

1. The cross-sectional area is \(A(x)=e^{-2x^2}\). 2. \(V=\int_0^2e^{-2x^2}\,\mathrm{d}x\). 3. Numerical evaluation gives \(V\approx0.627\) cubic units.

Answer

\(\approx0.627\) cubic units

Volumes with cross-sections 13

Two solids share the same base and use the same base width \(w(x)\) as a side length. One has square cross sections; the other has equilateral-triangle cross sections. If the square-cross-section volume is \(40\), find the other volume.

Hints

- Compare shape area formulas using the same side length. - Carry the constant ratio from slices to the entire solid.

Solution

1. Square area is \(w(x)^2\). 2. Equilateral-triangle area is \(\frac{\sqrt3}{4}w(x)^2\). 3. The same constant ratio holds after integration. 4. The triangle-cross-section volume is \(\frac{\sqrt3}{4}(40)=10\sqrt3\).

Answer

\(10\sqrt3\) cubic units

Volumes with cross-sections 14

A solid has isosceles-triangle cross sections perpendicular to the x-axis. At each \(x\in[0,3]\), both the triangle's base and altitude equal \(3-x\). Find the volume.

Hints

- Use the two equal but perpendicular triangle dimensions. - The one-half factor remains part of every cross-sectional area.

Solution

1. The area is \(A(x)=\frac12(3-x)(3-x)=\frac12(3-x)^2\). 2. \(V=\frac12\int_0^3(3-x)^2\,\mathrm{d}x=\frac{9}{2}\).

Answer

\(\frac{9}{2}\) cubic units

Volumes with cross-sections 15

For a certain base region, \(\int_a^b w(x)^2\,\mathrm{d}x=12\), where \(w(x)\) is the base width. A designer needs volume \(\frac{3\pi}{2}\). Which cross-section choice produces that volume? A. Squares with side \(w\) B. Circles with diameter \(w\) C. Semicircles with diameter \(w\) D. Equilateral triangles with side \(w\)

Hints

- Factor each shape's area into a constant times the squared width. - Use the supplied integral once for each possible constant.

Solution

1. The shape factors multiplying \(w^2\) are \(1\), \(\frac{\pi}{4}\), \(\frac{\pi}{8}\), and \(\frac{\sqrt3}{4}\), respectively. 2. Multiplying by \(\int_a^b w(x)^2\,\mathrm{d}x=12\), option C gives \(12\cdot\frac{\pi}{8}=\frac{3\pi}{2}\). 3. Therefore, semicircles with diameter \(w\) meet the design requirement.

Answer

Choice C: semicircles with diameter \(w\).

Volumes with cross-sections 16

The graph shows the base of a solid. Cross sections perpendicular to the x-axis are equilateral triangles whose side length is the vertical width of the base. Write the cross-sectional area function \(A(x)\) and the volume integral. Do not evaluate the integral.

Hints

- Read the vertical distance between the two displayed curves. - That distance is the side length, not the area, of each equilateral triangle. - Apply the equilateral-triangle area formula before integrating.

Solution

1. On \(0\le x\le2\), the vertical width is \(s(x)=2x-x^2\). 2. An equilateral triangle with side \(s\) has area \(\frac{\sqrt3}{4}s^2\), so \(A(x)=\frac{\sqrt3}{4}(2x-x^2)^2\). 3. The volume setup is \(V=\int_0^2\frac{\sqrt3}{4}(2x-x^2)^2\,\mathrm{d}x\).

Answer

\(A(x)=\frac{\sqrt3}{4}(2x-x^2)^2\) \(V=\int_0^2\frac{\sqrt3}{4}(2x-x^2)^2\,\mathrm{d}x\)

Volumes with cross-sections 17

The cross sections of a solid perpendicular to the x-axis are squares. At position \(x\), each square's diagonal is \(x+2\) centimeters. For \(0\leq x\leq2\), write the cross-sectional area \(A(x)\) and an integral for the solid's volume. Do not evaluate the integral.

Hints

- The given length runs corner to corner, not along one side. - Relate the diagonal and side of a square with a right triangle. - Volume comes from adding cross-sectional areas along the x-axis.

Solution

1. A square with diagonal \(d\) has side length \(\frac{d}{\sqrt2}\), so its area is \(\frac{d^2}{2}\). 2. Here \(A(x)=\frac{(x+2)^2}{2}\,\text{cm}^2\). 3. Integrating the slice areas gives \(V=\int_0^2\frac{(x+2)^2}{2}\,\mathrm{d}x\), where \(x\) is in centimeters.

Answer

\(A(x)=\frac{(x+2)^2}{2}\,\text{cm}^2\). \(V=\int_0^2\frac{(x+2)^2}{2}\,\mathrm{d}x\).

Volumes with cross-sections 18

Each cross section of a solid is a rectangle. One side is \(3\,\text{cm}\) and its diagonal is \(5\,\text{cm}\). All cross sections are identical, and the solid is \(2\,\text{cm}\) long perpendicular to them. Find the volume.

Hints

- A rectangle's diagonal and two sides form a right triangle. - Find the missing side before calculating the slice's area. - With identical slices, multiply the area of one by the solid's length.

Solution

1. By the Pythagorean theorem, the other rectangle side is \(\sqrt{5^2-3^2}=4\,\text{cm}\). 2. The area of each cross section is \(3\cdot4=12\,\text{cm}^2\). 3. The volume is \(12\,\text{cm}^2\cdot2\,\text{cm}=24\,\text{cm}^3\).

Answer

\(24\,\text{cm}^3\)

Volumes with cross-sections 19

The cross sections of a solid perpendicular to the x-axis are trapezoids. At position \(x\), their parallel sides are \(x+1\) and \(x+3\) centimeters, and their height is \(2\,\text{cm}\). Find the solid's volume for \(0\leq x\leq2\), with \(x\) measured in centimeters.

Hints

- Use the trapezoid area formula for each cross section. - The two parallel sides both change with position. - Add the slice areas by integrating over the given interval.

Solution

1. Trapezoid area is half the sum of the parallel sides times the height. 2. Thus \(A(x)=\frac{1}{2}((x+1)+(x+3))\cdot2=2x+4\,\text{cm}^2\). 3. The volume is \(V=\int_0^2(2x+4)\,\mathrm{d}x=[x^2+4x]_0^2=12\,\text{cm}^3\).

Answer

\(12\,\text{cm}^3\)

Volumes with cross-sections 20

A cross section of a solid is a square with a diagonal of \(6\,\text{cm}\). a) Find its area. b) If a different square cross section has diagonal \(d\), write its area in terms of \(d\).

Hints

- A square's diagonal divides it into two right triangles. - How is its side length related to its diagonal? - Use the side length to express area.

Solution

1. A square of side length \(s\) has diagonal \(s\sqrt{2}\) by the Pythagorean theorem. 2. Its area is \(s^2=\frac{d^2}{2}\). 3. For diagonal \(6\,\text{cm}\), the area is \(\frac{6^2}{2}=18\,\text{cm}^2\).

Answer

a) \(18\,\text{cm}^2\) b) \(A=\frac{d^2}{2}\)

Volumes with cross-sections 21

A goblet-shaped container has circular horizontal cross sections of radius \(f(x)=\sqrt{3x}\) for \(0\le x\le10\), where \(x\) is height above the bottom and both measurements are in centimeters. The container holds \(150\,\text{mL}\) of liquid. Find the liquid height exactly and to the nearest hundredth of a centimeter.

Hints

- Use the liquid height as the upper limit of a disc-method volume integral. - How are milliliters related to cubic centimeters? - After integrating, solve the resulting equation for the positive height.

Solution

1. If the liquid reaches height \(h\), its volume is \(V(h)=\pi\int_0^h[f(x)]^2\,\text{d}x\). 2. Thus \(V(h)=\pi\int_0^h3x\,\text{d}x=\frac32\pi h^2\). 3. Since \(150\,\text{mL}=150\,\text{cm}^3\), solve \(\frac32\pi h^2=150\). 4. Therefore, \(h=\sqrt{\frac{100}{\pi}}=\frac{10}{\sqrt{\pi}}\,\text{cm}\approx5.64\,\text{cm}\).

Answer

\(\frac{10}{\sqrt{\pi}}\,\text{cm}\approx5.64\,\text{cm}\)

Volumes with cross-sections 22

The shaded base of a solid lies between \(y=\sqrt{x}\) and \(y=\frac{x}{2}\) for \(0\le x\le4\). Cross sections perpendicular to the x-axis are squares. Find the volume.

Hints

- Use the vertical distance between the two base curves as the square's side length. - Square that distance to obtain cross-sectional area. - Integrate the cross-sectional area over \(0\le x\le4\).

Solution

1. The side length of a square is the vertical width of the base: \(s(x)=\sqrt x-\frac{x}{2}\). 2. The cross-sectional area is \(A(x)=s(x)^2\). 3. Therefore, \(V=\int_0^4\left(\sqrt x-\frac{x}{2}\right)^2\,\mathrm{d}x=\frac{8}{15}\).

Answer

\(\frac{8}{15}\) cubic units

Volumes with cross-sections 23

The shaded base of a solid lies between \(y=3\) and \(y=x^2\). Each cross section perpendicular to the x-axis is a rectangle whose height is three times its base in the xy-plane. Find the volume.

Hints

- Find the x-interval from the intersections of the base curves. - Use \(3-x^2\) as the rectangle's base dimension. - Multiply the two rectangle dimensions before integrating.

Solution

1. The base width is \(b(x)=3-x^2\), and the curves intersect at \(x=\pm\sqrt3\). 2. The rectangle height is \(3b(x)\), so its area is \(A(x)=b(x)\cdot3b(x)=3(3-x^2)^2\). 3. Therefore, \(V=\int_{-\sqrt3}^{\sqrt3}3(3-x^2)^2\,\mathrm{d}x=\frac{144\sqrt3}{5}\).

Answer

\(\frac{144\sqrt3}{5}\) cubic units

Volumes with cross-sections 24

The shaded base of a solid is enclosed by \(y=x\) and \(y=x^2\) in the first quadrant. Cross sections perpendicular to the x-axis are equilateral triangles whose sides lie in the base. Find the exact volume.

Hints

- Use the vertical separation of the base curves as the triangle's side length. - Apply the equilateral-triangle area formula to that variable side. - Integrate from the two intersection x-values.

Solution

1. The curves intersect at \(x=0\) and \(x=1\), and the triangle side length is \(s(x)=x-x^2\). 2. The cross-sectional area is \(A(x)=\frac{\sqrt3}{4}s(x)^2\). 3. Therefore, \(V=\frac{\sqrt3}{4}\int_0^1(x-x^2)^2\,\mathrm{d}x=\frac{\sqrt3}{120}\).

Answer

\(\frac{\sqrt3}{120}\) cubic units

Volumes with cross-sections 25

The shaded base of a solid is bounded by \(y=4\) and \(y=x^2\). Cross sections perpendicular to the x-axis are semicircles whose diameters lie in the base. Find the volume.

Hints

- Use the vertical base width as the semicircle's diameter, not its radius. - Express semicircle area in terms of the diameter. - Integrate over the full interval from \(x=-2\) to \(x=2\).

Solution

1. The diameter is \(d(x)=4-x^2\) for \(-2\le x\le2\). 2. A semicircle with diameter \(d\) has area \(A=\frac{\pi}{8}d^2\). 3. Therefore, \(V=\frac{\pi}{8}\int_{-2}^{2}(4-x^2)^2\,\mathrm{d}x=\frac{64\pi}{15}\).

Answer

\(\frac{64\pi}{15}\) cubic units

Volumes with cross-sections 26

For \(-1\le x\le1\), a base region has vertical width \(w(x)=1-x^2\). Cross sections perpendicular to the x-axis are regular hexagons whose long diagonals equal \(w(x)\). The long diagonal of a regular hexagon is twice its side length. a) Write the cross-sectional area \(A(x)\). b) Find the exact volume.

Hints

- Convert the long diagonal into the side length of the regular hexagon. - The cross-sectional area depends on the square of the base width. - Use symmetry when evaluating the volume integral.

Solution

1. If the long diagonal is \(w(x)\), then the hexagon side length is \(s(x)=\frac{w(x)}{2}\). 2. Using \(A=\frac{3\sqrt3}{2}s^2\), the cross-sectional area is \(A(x)=\frac{3\sqrt3}{8}(1-x^2)^2\). 3. The volume is \(V=\frac{3\sqrt3}{8}\int_{-1}^{1}(1-x^2)^2\,\mathrm{d}x\). 4. Since \(\int_{-1}^{1}(1-x^2)^2\,\mathrm{d}x=\frac{16}{15}\), \(V=\frac{2\sqrt3}{5}\).

Answer

a) \(A(x)=\frac{3\sqrt3}{8}(1-x^2)^2\) b) \(\frac{2\sqrt3}{5}\) cubic units

Volumes with cross-sections 27

The shaded base is enclosed by \(x=2y\) and \(x=y^2\). Cross sections perpendicular to the y-axis are squares. Find the volume.

Hints

- Cross sections perpendicular to the y-axis use horizontal base widths. - Subtract \(x=y^2\) from \(x=2y\) to obtain the square's side. - Square the side length before integrating with respect to \(y\).

Solution

1. The curves intersect at \(y=0\) and \(y=2\). 2. The square side is the horizontal width \(s(y)=2y-y^2\). 3. Therefore, \(V=\int_0^2(2y-y^2)^2\,\mathrm{d}y=\frac{16}{15}\).

Answer

\(\frac{16}{15}\) cubic units

Volumes with cross-sections 28

The shaded base lies between \(x=y^2\) and \(x=4\) for \(-2\le y\le2\). Cross sections perpendicular to the y-axis are rectangles. Each rectangle has base equal to the horizontal width of the region and height \(y+3\). Find the volume.

Hints

- Use the horizontal distance from \(x=y^2\) to \(x=4\) as one rectangle dimension. - Use the separately stated height \(y+3\) as the other dimension. - Integrate the product over \(-2\le y\le2\).

Solution

1. The rectangle base is \(b(y)=4-y^2\), and its other dimension is \(y+3\). 2. Thus, the cross-sectional area is \(A(y)=(4-y^2)(y+3)\). 3. Therefore, \(V=\int_{-2}^{2}(4-y^2)(y+3)\,\mathrm{d}y=32\).

Answer

\(32\) cubic units

Volumes with cross-sections 29

For \(0\le x\le3\), the shaded base lies between \(y=x\) and \(y=3-x\). Cross sections perpendicular to the x-axis are squares. Find the volume, explaining why the boundary switch does not require splitting the final integral.

Hints

- A side length is a distance, so use an absolute value when the upper line changes. - Squaring an absolute value gives the same result as squaring its expression. - Locate the line intersection to explain the change in geometric order.

Solution

1. The vertical distance between the lines is \(s(x)=|(3-x)-x|=|3-2x|\). 2. A square cross section has area \(A(x)=s(x)^2=|3-2x|^2=(3-2x)^2\). 3. Squaring removes the sign, so one area formula works across the intersection at \(x=\frac{3}{2}\). 4. Therefore, \(V=\int_0^3(3-2x)^2\,\mathrm{d}x=9\).

Answer

\(9\) cubic units

Volumes with cross-sections 30

The shaded base region is shown. Cross sections perpendicular to the x-axis are squares. Find the volume.

Hints

- Use two grid points on each line to determine its equation. - Subtract the lower y-value from the upper y-value to get the square side. - Square the side length and integrate to the intersection.

Solution

1. From the graph, the upper boundary is \(y=3-\frac{1}{2}x\) and the lower boundary is \(y=\frac{1}{2}x\). Thus, the square side length is \(s(x)=3-x\). 2. The boundaries meet at \(x=3\). 3. Therefore, \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\).

Answer

\(9\) cubic units

Volumes with cross-sections 31

For \(0\le x\le1\), the vertical width of a base region is \(w(x)=kx(1-x)\), where \(k>0\). Cross sections perpendicular to the x-axis are squares. If the volume is \(\frac{6}{5}\), find \(k\).

Hints

- The square side length is the given vertical width. - Squaring the width makes the volume depend on \(k^2\). - Use \(k>0\) after solving the volume equation.

Solution

1. The square area is \(A(x)=k^2x^2(1-x)^2\). 2. Therefore, \(V=k^2\int_0^1x^2(1-x)^2\,\mathrm{d}x=\frac{k^2}{30}\). 3. Set \(\frac{k^2}{30}=\frac{6}{5}\). Then \(k^2=36\), and \(k=6\) because \(k>0\).

Answer

\(k=6\)

Volumes with cross-sections 32

A solid has cross sections perpendicular to the x-axis that are semicircles with diameter \(d(x)=x\), from \(x=0\) to \(x=b\). The volume is \(9\pi\). Find \(b\).

Hints

- Express the volume in terms of the unknown endpoint. - Cancel the common geometric constant before solving.

Solution

1. The cross-sectional area is \(A(x)=\frac{\pi}{8}x^2\). 2. \(V=\frac{\pi}{8}\int_0^b x^2\,\mathrm{d}x=\frac{\pi b^3}{24}\). 3. Set \(\frac{\pi b^3}{24}=9\pi\), giving \(b^3=216\) and \(b=6\).

Answer

\(b=6\)

Volumes with cross-sections 33

A model solid has base region between \(y=2-\frac{x}{2}\) and \(y=0\) for \(0\le x\le4\). Cross sections perpendicular to the x-axis are squares. One coordinate unit represents \(2\,\text{cm}\) in the actual object. Find the actual volume.

Hints

- What is the vertical distance between the two base boundaries at position \(x\)? - Compute the model volume before applying the physical scale factor. - How does volume scale when every linear dimension is multiplied by \(2\)?

Solution

1. In coordinate units, \(V_m=\int_0^4\left(2-\frac{x}{2}\right)^2\,\mathrm{d}x=\frac{16}{3}\). 2. A linear scale factor of \(2\) multiplies volume by \(2^3=8\). 3. Therefore, the actual volume is \(8V_m=\frac{128}{3}\,\text{cm}^3\).

Answer

\(\frac{128}{3}\,\text{cm}^3\)

Volumes with cross-sections 34

A base lies between \(y=1\) and \(y=x^2\). Cross sections perpendicular to the x-axis are right triangles. One leg equals the base width, and the other leg is twice that width. Find the volume.

Hints

- How do the two base curves determine the vertical width at a given \(x\)? - Express both triangle legs in terms of that width before finding area. - Volume is the integral of the resulting cross-sectional area.

Solution

1. The base width is \(w(x)=1-x^2\) for \(-1\le x\le1\). 2. The triangle area is \(A(x)=\frac{1}{2}w(x)[2w(x)]=w(x)^2\). 3. Therefore, \(V=\int_{-1}^{1}(1-x^2)^2\,\mathrm{d}x=\frac{16}{15}\).

Answer

\(\frac{16}{15}\) cubic units

Volumes with cross-sections 35

A base lies between \(y=\sin x\) and the x-axis on \([0,\pi]\). Cross sections perpendicular to the x-axis are semicircles whose diameters lie in the base. Find the exact volume.

Hints

- The vertical base width is the semicircle's diameter, not its radius. - Write semicircle area in terms of diameter before integrating. - What is \(\int_0^\pi\sin^2x\,\mathrm{d}x\)?

Solution

1. The diameter is \(d(x)=\sin x\). 2. The cross-sectional area is \(A(x)=\frac{\pi}{8}\sin^2x\). 3. Therefore, \(V=\frac{\pi}{8}\int_0^\pi\sin^2x\,\mathrm{d}x=\frac{\pi^2}{16}\).

Answer

\(\frac{\pi^2}{16}\) cubic units

Volumes with cross-sections 36

The base is bounded by \(y=4\) and \(y=x^2\). Perpendicular cross sections are squares whose diagonals, rather than sides, lie in the base. The diagram shows one cross section. Find the volume.

Hints

- Use the base width as the square's diagonal, not its side. - Express square area in terms of diagonal length. - Integrate over the intersections \(x=-2\) and \(x=2\).

Solution

1. The square diagonal is the vertical base width \(d(x)=4-x^2\). 2. A square with diagonal \(d\) has area \(A=\frac{d^2}{2}\). 3. Therefore, \(V=\frac{1}{2}\int_{-2}^{2}(4-x^2)^2\,\mathrm{d}x=\frac{256}{15}\).

Answer

\(\frac{256}{15}\) cubic units

Volumes with cross-sections 37

A base is bounded by \(x=0\), \(x=4-y\), \(y=0\), and \(y=4\). Cross sections perpendicular to the y-axis are circles whose circumferences equal the horizontal widths of the base. Find the volume.

Hints

- What is the horizontal width of the base at height \(y\)? - Convert circumference to radius before calculating circle area. - Integrate with respect to \(y\) over the full height of the base.

Solution

1. At height \(y\), the horizontal width is \(w(y)=4-y\). 2. If a circle's circumference is \(w(y)\), then its radius is \(r(y)=\frac{4-y}{2\pi}\). 3. The cross-sectional area is \(A(y)=\pi[r(y)]^2=\frac{(4-y)^2}{4\pi}\). 4. Therefore, \(V=\frac{1}{4\pi}\int_0^4(4-y)^2\,\mathrm{d}y=\frac{16}{3\pi}\).

Answer

\(\frac{16}{3\pi}\) cubic units

Volumes with cross-sections 38

A base has vertical width \(x\) for \(0\le x\le3\). Each cross section perpendicular to the x-axis is a rectangle with one side equal to that width. The representative cross section shows its diagonal length. Find the volume.

Hints

- Read the fixed diagonal length from the representative cross section. - How can the diagonal and the two rectangle side lengths be related? - Build the rectangular area function before integrating.

Solution

1. One side of a cross-sectional rectangle is \(x\). The diagram gives diagonal length \(5\). If the other side is \(h(x)\), then \(x^2+[h(x)]^2=25\), so \(h(x)=\sqrt{25-x^2}\). 2. The cross-sectional area is \(A(x)=x\sqrt{25-x^2}\). 3. The volume is \(V=\int_0^3x\sqrt{25-x^2}\,\mathrm{d}x\). 4. Evaluating gives \(V=\frac{1}{3}(125-64)=\frac{61}{3}\).

Answer

\(\frac{61}{3}\) cubic units

Volumes with cross-sections 39

In the first quadrant, the shaded base lies between \(y=x\) and \(y=x^2\). Cross sections perpendicular to the y-axis are semicircles whose diameters lie in the base. Find the volume.

Hints

- Rewrite both base curves as x-functions of y. - Use the horizontal base width as the semicircle diameter. - Express semicircle area in terms of diameter before integrating.

Solution

1. Rewrite the boundaries as \(x=y\) and \(x=\sqrt y\). 2. The horizontal diameter is \(d(y)=\sqrt y-y\) for \(0\le y\le1\). 3. Therefore, \(V=\frac{\pi}{8}\int_0^1(\sqrt y-y)^2\,\mathrm{d}y=\frac{\pi}{240}\).

Answer

\(\frac{\pi}{240}\) cubic units

Volumes with cross-sections 40

The shaded base of a solid is the disk \(x^2+y^2\le9\). Cross sections perpendicular to the x-axis are squares. Find the volume.

Hints

- Solve the circle equation for the upper and lower y-values. - Use the full vertical chord, not one radius, as the square side. - Square the chord length before integrating across the disk.

Solution

1. At position \(x\), the disk's vertical chord has length \(s(x)=2\sqrt{9-x^2}\). 2. The square area is \(A(x)=4(9-x^2)\). 3. Therefore, \(V=\int_{-3}^{3}4(9-x^2)\,\mathrm{d}x=144\).

Answer

\(144\) cubic units

Volumes with cross-sections 41

The shaded base of a solid is the disk \(x^2+y^2\le4\). Cross sections perpendicular to the x-axis are semicircles whose diameters are vertical chords of the disk. Find the volume.

Hints

- Solve the circle equation for both vertical endpoints of a chord. - Use the full chord as the semicircle diameter. - Simplify the squared radical before integrating.

Solution

1. The chord length is \(d(x)=2\sqrt{4-x^2}\). 2. The semicircle area is \(A(x)=\frac{\pi}{8}d(x)^2=\frac{\pi}{2}(4-x^2)\). 3. Therefore, \(V=\frac{\pi}{2}\int_{-2}^{2}(4-x^2)\,\mathrm{d}x=\frac{16\pi}{3}\).

Answer

\(\frac{16\pi}{3}\) cubic units

Volumes with cross-sections 42

A triangular base has vertices \((0,0)\), \((3,0)\), and \((0,3)\). Cross sections perpendicular to the x-axis are squares. a) Write the cross-sectional area function \(A(x)\). b) Use \(V=\int A(x)\,\mathrm{d}x\) to find the exact volume. c) Identify the standard solid formed and verify the same volume with its geometric volume formula.

Hints

- Determine the vertical width of the triangular base at a general x-value. - That width is a square side length, so how is \(A(x)\) related to it? - After integrating, compare the dimensions with a familiar pyramid formula.

Solution

1. The slanted boundary through \((0,3)\) and \((3,0)\) is \(y=3-x\), so each square has side length \(s(x)=3-x\) for \(0\le x\le3\). 2. Therefore, \(A(x)=(3-x)^2\). 3. Using cross-sectional integration, \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\). 4. The linearly shrinking square sections form a square pyramid with base side \(3\) and height \(3\). Its geometric formula gives \(\frac{1}{3}(3^2)(3)=9\), confirming the integral.

Answer

a) \(A(x)=(3-x)^2\) b) \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\) cubic units c) The solid is a square pyramid, and \(\frac{1}{3}(3^2)(3)=9\) cubic units.

Volumes with cross-sections 43

A sculptural beam extends from \(x=0\) to \(x=\pi\), where \(x\) is measured in feet. Its base has vertical width \(w(x)=2+\sin x\) feet. Cross sections perpendicular to the x-axis are equilateral triangles with side \(w(x)\). a) Write the cross-sectional area \(A(x)\). b) Find the exact volume of the beam.

Hints

- What is the area formula for an equilateral triangle in terms of its side length? - Substitute the given width function before forming the volume integral. - How can \(\sin^2x\) be integrated over \([0,\pi]\)?

Solution

1. The area of an equilateral triangle with side \(w\) is \(\frac{\sqrt3}{4}w^2\), so \(A(x)=\frac{\sqrt3}{4}(2+\sin x)^2\). 2. The volume is \(V=\frac{\sqrt3}{4}\int_0^\pi(2+\sin x)^2\,\mathrm{d}x\). 3. Expand and integrate: \(\int_0^\pi(4+4\sin x+\sin^2x)\,\mathrm{d}x=4\pi+8+\frac{\pi}{2}=8+\frac{9\pi}{2}\). 4. Therefore, \(V=2\sqrt3+\frac{9\pi\sqrt3}{8}\,\text{ft}^3\).

Answer

a) \(A(x)=\frac{\sqrt3}{4}(2+\sin x)^2\) b) \(2\sqrt3+\frac{9\pi\sqrt3}{8}\,\text{ft}^3\)

Volumes with cross-sections 44

Square cross sections of a solid have side lengths measured at several x-values. <table><tr><td>\(x\)</td><td>\(0\)</td><td>\(1\)</td><td>\(3\)</td><td>\(5\)</td></tr><tr><td>side \(s(x)\)</td><td>\(0\)</td><td>\(2\)</td><td>\(3\)</td><td>\(1\)</td></tr></table> Convert the side lengths to cross-sectional areas, then use the trapezoidal rule on those area values to estimate the volume.

Hints

- What area corresponds to each measured square side length? - Use the actual interval widths \(1\), \(2\), and \(2\). - Apply the trapezoidal rule to cross-sectional area, not directly to side length.

Solution

1. Square the side data to obtain cross-sectional areas \(0\), \(4\), \(9\), and \(1\). 2. Apply trapezoids to \(A(x)\), not to \(s(x)\): \(\frac{1}{2}(0+4)+\frac{2}{2}(4+9)+\frac{2}{2}(9+1)=2+13+10\). 3. Therefore, \(V\approx25\) cubic units.

Answer

\(V\approx25\) cubic units

Volumes with cross-sections 45

Square cross sections are measured at three x-values. <table><tr><td>\(x\)</td><td>\(0\)</td><td>\(2\)</td><td>\(4\)</td></tr><tr><td>side length</td><td>\(1\)</td><td>\(m\)</td><td>\(3\)</td></tr></table> A trapezoidal estimate using the cross-sectional areas gives a volume of \(20\) cubic units. Find \(m>0\).

Hints

- Convert the side lengths to square areas before using the trapezoidal rule. - Each subinterval has width \(2\). - Use the positivity condition after solving for \(m^2\).

Solution

1. The cross-sectional area values are \(1\), \(m^2\), and \(9\). 2. With subinterval width \(2\), the trapezoidal estimate is \((1+m^2)+(m^2+9)=10+2m^2\). 3. Set \(10+2m^2=20\). Then \(m^2=5\), so \(m=\sqrt5\) because \(m>0\).

Answer

\(m=\sqrt5\)

Volumes with cross-sections 46

A base lies between \(y=\cos x\) and \(y=\sin x\) for \(0\le x\le\frac{\pi}{4}\). Cross sections perpendicular to the x-axis are squares. Find the exact volume.

Hints

- Which trigonometric curve is above the other on the stated interval? - The square side is the vertical distance between the curves. - Simplify the squared difference before integrating.

Solution

1. On the interval, \(\cos x\ge\sin x\), so the square side length is \(s(x)=\cos x-\sin x\). 2. The cross-sectional area is \(A(x)=(\cos x-\sin x)^2=1-\sin(2x)\). 3. Therefore, \(V=\int_0^{\pi/4}[1-\sin(2x)]\,\mathrm{d}x=\frac{\pi}{4}-\frac{1}{2}\).

Answer

\(\frac{\pi}{4}-\frac{1}{2}\) cubic units

Volumes with cross-sections 47

A base lies between \(y=e^x\) and \(y=1+x\) on \([0,1]\). Cross sections perpendicular to the x-axis are squares. Find the volume to four decimal places.

Hints

- Determine which base curve is above on \([0,1]\). - Use their vertical separation as the square side length. - Set up the exact integral before using numerical technology.

Solution

1. Since \(e^x\ge1+x\), the square side length is \(s(x)=e^x-1-x\). 2. The cross-sectional area is \(A(x)=(e^x-1-x)^2\), so \(V=\int_0^1(e^x-1-x)^2\,\mathrm{d}x\). 3. Numerical evaluation gives \(V\approx0.0913\) cubic units.

Answer

\(\approx0.0913\) cubic units

Volumes with cross-sections 48

The graph shows the upper and lower boundaries of a base region. Cross sections perpendicular to the x-axis are squares whose diagonals lie in the base. Write and evaluate the volume integral.

Hints

- Read the square diagonal as the vertical separation in the base. - Express a square's area in terms of its diagonal. - Use symmetry to remove the absolute value before integrating.

Solution

1. The upper boundary is \(y=2\) and the lower boundary is \(y=|x|\), so the square diagonal is \(d(x)=2-|x|\) on \([-2,2]\). 2. A square with diagonal \(d\) has area \(\frac{1}{2}d^2\). Thus, \(A(x)=\frac{1}{2}(2-|x|)^2\). 3. By symmetry, \(V=2\int_0^2\frac{1}{2}(2-x)^2\,\mathrm{d}x=\frac{8}{3}\) cubic units.

Answer

\(\frac{8}{3}\) cubic units

Volumes with cross-sections 49

For \(0\le x\le4\), each cross section is a rectangle. One side has length \(x\), and the rectangle's perimeter is always \(12\). Find the volume.

Hints

- Use the perimeter condition to recover the missing rectangle dimension. - Verify both dimensions stay nonnegative on the interval.

Solution

1. If the other side is \(h(x)\), then \(2[x+h(x)]=12\), so \(h(x)=6-x\). 2. The area is \(A(x)=x(6-x)\). 3. \(V=\int_0^4x(6-x)\,\mathrm{d}x=\frac{80}{3}\).

Answer

\(\frac{80}{3}\) cubic units

Volumes with cross-sections 50

A base lies between \(y=\ln x\) and the x-axis for \(1\le x\le e\). Cross sections perpendicular to the x-axis are squares. Find the exact volume.

Hints

- What is the vertical distance from the x-axis to the logarithmic curve? - Square that distance to form each square cross-sectional area. - Evaluate the logarithmic antiderivative at \(x=1\) and \(x=e\).

Solution

1. The square side length is the vertical distance from the x-axis to the curve: \(s(x)=\ln x\). 2. The cross-sectional area is \(A(x)=(\ln x)^2\), so \(V=\int_1^e(\ln x)^2\,\mathrm{d}x\). 3. An antiderivative is \(x[(\ln x)^2-2\ln x+2]\). Therefore, \(V=e-2\) cubic units.

Answer

\(e-2\) cubic units

Volumes with cross-sections 51

A pedestrian tunnel is \(10\,\text{m}\) long. Cross sections perpendicular to the x-axis are equilateral triangles. The side length decreases linearly from \(6\,\text{m}\) at \(x=0\) to \(2\,\text{m}\) at \(x=10\). a) Find the side-length function \(s(x)\). b) Find the exact volume of the tunnel.

Hints

- Build a linear function from the two endpoint side lengths. - Convert side length into equilateral-triangle area before integrating. - Check that your side function gives the stated endpoint values.

Solution

1. The side length has slope \(\frac{2-6}{10}=-\frac{2}{5}\), so \(s(x)=6-\frac{2}{5}x\). 2. The cross-sectional area is \(A(x)=\frac{\sqrt3}{4}\left(6-\frac{2}{5}x\right)^2\). 3. The volume is \(V=\frac{\sqrt3}{4}\int_0^{10}\left(6-\frac{2}{5}x\right)^2\,\mathrm{d}x\). 4. The integral equals \(\frac{520}{3}\), so \(V=\frac{130\sqrt3}{3}\,\text{m}^3\).

Answer

a) \(s(x)=6-\frac{2}{5}x\) b) \(\frac{130\sqrt3}{3}\,\text{m}^3\)

Volumes with cross-sections 52

A base has vertical width \(w(x)=1-x\) for \(0\le x\le1\). Cross sections are rectangles whose perpendicular height is \((1+x)w(x)\). Find the volume.

Hints

- The aspect ratio changes with x, so keep its factor inside the integral. - Both rectangle dimensions include the base width.

Solution

1. The cross-sectional area is \(A(x)=w(x)[(1+x)w(x)]=(1+x)(1-x)^2\). 2. \(V=\int_0^1(1+x)(1-x)^2\,\mathrm{d}x=\frac{5}{12}\).

Answer

\(\frac{5}{12}\) cubic units

Volumes with cross-sections 53

A glass container has circular horizontal cross sections. Its bottom is at height \(x=-6\), and its rim is at \(x=0\). Its inside radius at height \(x\) is \(f(x)=\sqrt{\frac{12}{\pi}\cos\left(\frac{\pi x}{12}\right)}\). All lengths are in inches. a) Find the water volume when the water is \(3\,\text{in.}\) deep. Give an exact value and a value rounded to the nearest hundredth of a cubic inch. b) Adding \(4\,\text{in.}^3\) of water raises the water level from a depth \(h\) by exactly \(1\,\text{in.}\). Write an equation that can be used to find \(h\), where \(0\le h\le5\).

Hints

- Depth is measured upward from the bottom. How does that differ from the height coordinate \(x\)? - What area does a thin horizontal layer of water have? - In part b), what change in the volume function corresponds to one more inch of depth?

Solution

1. For water depth \(h\), \(V(h)=\pi\int_{-6}^{-6+h}[f(x)]^2\,\text{d}x=12\int_{-6}^{-6+h}\cos\left(\frac{\pi x}{12}\right)\,\text{d}x\). 2. Therefore \(V(h)=\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-6)}{12}\right)+1\right)\). 3. At \(h=3\), \(V(3)=\frac{144}{\pi}\left(1-\frac{\sqrt2}{2}\right)\,\text{in.}^3\approx13.43\,\text{in.}^3\). 4. The added water is the change \(V(h+1)-V(h)=4\). 5. Substituting gives \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\).

Answer

a) \(\frac{144}{\pi}\left(1-\frac{\sqrt2}{2}\right)\,\text{in.}^3\approx13.43\,\text{in.}^3\) b) \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\)

Volumes with cross-sections 54

For \(0\le x\le2\), the accumulated volume of a solid is \(V(x)=x^3-3x^2+4x\). Cross sections perpendicular to the x-axis are squares. Find the square side-length function and the location and value of its shortest side.

Hints

- Differentiate accumulated volume to obtain cross-sectional area. - Take the positive square root of the square's area to obtain its side length. - Minimize the quadratic area expression before taking its square root.

Solution

1. The cross-sectional area is \(V'(x)=3x^2-6x+4\). 2. Since the sections are squares, \(s(x)=\sqrt{3x^2-6x+4}\). 3. Rewrite the area as \(3(x-1)^2+1\), which is minimized at \(x=1\). 4. The shortest side has length \(s(1)=1\).

Answer

\(s(x)=\sqrt{3x^2-6x+4}\); the shortest side is \(1\) unit at \(x=1\).

Volumes with cross-sections 55

The shaded base is bounded by \(x=y^2\), \(x=4\), \(y=0\), and \(y=2\). Cross sections perpendicular to the y-axis are squares. Find the height \(c\) that divides the solid into two equal-volume parts. Round to three decimal places.

Hints

- Use the horizontal base width as the square side. - Calculate half of the total volume before forming the cut-height equation. - Solve the accumulation equation numerically within \(0<c<2\).

Solution

1. At height \(y\), the square side is \(4-y^2\), so \(A(y)=(4-y^2)^2\). 2. The total volume is \(\int_0^2(4-y^2)^2\,\mathrm{d}y=\frac{256}{15}\). 3. Equal volumes require \(\int_0^c(4-y^2)^2\,\mathrm{d}y=\frac{128}{15}\). 4. Numerical solution on \((0,2)\) gives \(c\approx0.562\).

Answer

\(c\approx0.562\)

Volumes with cross-sections 56

The graph shows the base of a solid. Cross sections perpendicular to the x-axis are squares whose side lengths are the vertical widths of the base. Determine the two width functions required by the graph, then find the exact volume of the solid.

Hints

- Identify where the lower boundary changes from the slanted line to the x-axis. - On each interval, write vertical width as upper boundary minus lower boundary. - Square each width before integrating, then add the two volume contributions.

Solution

1. For \(-1\le x\le0\), the upper boundary is \(y=4-x^2\) and the lower boundary is \(y=-3x\), so the square side is \(s_1(x)=4+3x-x^2\). 2. For \(0\le x\le2\), the lower boundary is the x-axis, so the side is \(s_2(x)=4-x^2\). 3. Therefore, \(V=\int_{-1}^{0}(4+3x-x^2)^2\,\mathrm{d}x+\int_{0}^{2}(4-x^2)^2\,\mathrm{d}x\). 4. The two integrals are \(\frac{181}{30}\) and \(\frac{256}{15}\), respectively, so \(V=\frac{231}{10}\) cubic units.

Answer

\(s_1(x)=4+3x-x^2\) for \(-1\le x\le0\) \(s_2(x)=4-x^2\) for \(0\le x\le2\) \(V=\frac{231}{10}\) cubic units

Volumes with cross-sections 57

The solid shown is a square pyramid. Let \(z\) measure distance from the apex toward the base along the height. Cross sections perpendicular to the height are squares. Using the dimensions shown, derive the cross-sectional area \(A(z)\), integrate it to find the pyramid's volume, and verify that the result agrees with \(V=\frac{1}{3}Bh\).

Hints

- Compare a cross-sectional square with the base square using similarity from the apex. - How does side length scale with distance along the pyramid's height? - Integrate cross-sectional area from the apex to the base, then compare with the standard pyramid formula.

Solution

1. The full height is \(9\,\text{cm}\) and the base side length is \(6\,\text{cm}\). By similarity, a square at distance \(z\) from the apex has side length \(s(z)=\frac{6}{9}z=\frac{2}{3}z\). 2. Thus, \(A(z)=s(z)^2=\frac{4}{9}z^2\). 3. The cross-sectional integral gives \(V=\int_0^9\frac{4}{9}z^2\,\mathrm{d}z=108\,\text{cm}^3\). 4. The base area is \(B=6^2=36\,\text{cm}^2\), and \(\frac{1}{3}Bh=\frac{1}{3}(36)(9)=108\,\text{cm}^3\), so the formulas agree.

Answer

\(A(z)=\frac{4}{9}z^2\), and \(V=108\,\text{cm}^3\).

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