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Volumes with cross-sections

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54949112
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

\(15\,\text{in}^3\)
54950312
A solid extends from \(x=0\) to \(x=2\). Each perpendicular cross section is a circle whose radius is \(2-x\). A representative cross section is shown. Find the volume.
Figure for problem 549503

Hints

- Use the given dimension directly as the radius. - Volume is the accumulation of the circular areas.

Solution

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

Answer

\(\frac{8\pi}{3}\) cubic units
54949012
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.
Figure for problem 549490

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
54949512
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.
54949612
A base has vertical width \(w(x)=2\sin x\) on \([0,\pi]\), and perpendicular cross sections are squares. The diagram shows one square cross section. A student writes \(V=\int_0^\pi4w(x)\,\mathrm{d}x\). Identify the error and find the correct volume.
Figure for problem 549496

Hints

- Volume accumulates two-dimensional cross-sectional area. - The base width is the side length of each square. - Square \(w(x)\) before integrating over the 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

The student integrated perimeter. The correct volume is \(2\pi\) cubic units.
54949912
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
54950012
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
54951512
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
54951912
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
54952112
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
54952512
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 \(w\).
54948212
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.
Figure for problem 549482

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
54948312
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.
Figure for problem 549483

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
54948412
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.
Figure for problem 549484

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
54948512
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.
Figure for problem 549485

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
54948612
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.
Figure for problem 549486

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
54948712
The shaded base is enclosed by \(x=2y\) and \(x=y^2\). Cross sections perpendicular to the y-axis are squares. Find the volume.
Figure for problem 549487

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
54948812
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.
Figure for problem 549488

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
54948912
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.
Figure for problem 549489

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
54949212
The shaded base region is shown. Cross sections perpendicular to the x-axis are squares. Find the volume.
Figure for problem 549492

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
54949312
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

\(k=6\)
54949412
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

\(b=6\)
54949712
The shaded base of a model solid lies 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.
Figure for problem 549497

Hints

- Use the displayed vertical base width as the square side in coordinate units. - Calculate the model volume before applying the physical scale. - A three-dimensional quantity scales by the cube of the linear factor.

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

\(\frac{128}{3}\,\text{cm}^3\)
54949812
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 \(x=1\).
54950112
The shaded 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.
Figure for problem 549501

Hints

- Use the vertical distance between the two base curves as one triangle leg. - Write the second leg as twice that distance. - Simplify the triangle-area formula before integrating.

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
54950212
The shaded 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.
Figure for problem 549502

Hints

- Use the vertical base width as a diameter, not as a radius. - Express semicircle area in terms of its diameter. - Evaluate the sine-squared integral over one full arch.

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
54950412
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.
Figure for problem 549504

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
54950512
The shaded base of a solid 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.
Figure for problem 549505

Hints

- Use the horizontal base width at height \(y\). - Convert circumference to radius before calculating circle area. - Integrate the circular areas from \(y=0\) to \(y=4\).

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
54950612
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 and diagonal length \(5\). A representative cross section is shown. Find the volume.
Figure for problem 549506

Hints

- Use the fixed diagonal to relate the two side lengths of a cross section. - Choose the positive side length that is consistent with a rectangle. - Build the area function before accumulating volume.

Solution

1. One side of a cross-sectional rectangle is \(x\). 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 \(V=\frac13(125-64)=\frac{61}{3}\).

Answer

\(\frac{61}{3}\) cubic units
54950712
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.
Figure for problem 549507

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
54950812
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.
Figure for problem 549508

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

\(c\approx0.562\)
54950912
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.
Figure for problem 549509

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
54951012
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.
Figure for problem 549510

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
54951112
The shaded triangular base has vertices \((0,0)\), \((3,0)\), and \((0,3)\). Cross sections perpendicular to the x-axis are squares. Find the volume, and identify the standard solid formed by these dimensions.
Figure for problem 549511

Hints

- Use the vertical width of the triangular base as the square side. - Notice how the square side decreases linearly to \(0\). - Compare the resulting dimensions with the formula for a square pyramid.

Solution

1. The slanted boundary is \(y=3-x\), so the square side is \(s(x)=3-x\). 2. Therefore, \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\). 3. The linearly shrinking square sections form a square pyramid with base side \(3\) and height \(3\). Its formula \(\frac{1}{3}(3^2)(3)=9\) agrees.

Answer

The volume is \(9\) cubic units; the solid is a square pyramid.
54951212
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.
Figure for problem 549512

Hints

- Use the equilateral-triangle area formula with the full base width as the side. - Expand the squared trigonometric expression before integrating. - Use the standard integral of \(\sin^2x\) 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, \(V=2\sqrt3+\frac{9\pi\sqrt3}{8}\).

Answer

a) \(A(x)=\frac{\sqrt3}{4}(2+\sin x)^2\) b) \(2\sqrt3+\frac{9\pi\sqrt3}{8}\,\text{ft}^3\)
54951312
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> The graph shows the corresponding cross-sectional area values. Use the trapezoidal rule on those area values to estimate the volume.
Figure for problem 549513

Hints

- Convert every measured side length into a square area first. - Use the actual interval widths \(1\), \(2\), and \(2\). - Add the three trapezoidal approximations to cross-sectional area.

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
54951412
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

\(m=\sqrt5\)
54951612
The 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.
Figure for problem 549516

Hints

- Determine which trigonometric curve is above on the interval. - The square side is the vertical distance between the curves. - Simplify the squared difference with a double-angle identity 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
54951712
The 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.
Figure for problem 549517

Hints

- Confirm which base curve is above on \([0,1]\). - The square side is the vertical separation between the curves. - 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
54951812
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.
Figure for problem 549518

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
54952012
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
54952212
The 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.
Figure for problem 549522

Hints

- Use the vertical distance from the x-axis as the square side length. - Square the width before integrating. - 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
54952312
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.
Figure for problem 549523

Hints

- Build the linear side-length function from its two endpoint values. - Convert side length into equilateral-triangle area before integrating. - Check that the side function gives \(6\) and \(2\) at the stated endpoints.

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) \(\frac{130\sqrt3}{3}\,\text{m}^3\)
54952412
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

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