Volumes with Known Cross Sections

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Key Ideas Easy Medium Hard

\(V=\int_a^b A(x)\,dx\)

What to Remember

The slice gives the area function

If cross sections perpendicular to the x-axis have area \(A(x)\), then

\(V=\int_a^b A(x)\,dx\)

Match the shape to A(x)

Square of side \(s\): \(s^2\). Semicircle of radius \(r\): \(\dfrac{1}{2}\pi r^2\).

Equilateral triangle of side \(s\): \(\dfrac{\sqrt{3}}{4}s^2\). Isosceles right triangle with leg \(s\): \(\dfrac{1}{2}s^2\).

The base supplies the side or the diameter

Read whether the segment in the base is a side, a leg, or a diameter. A diameter is twice the radius.

Perpendicular to the y-axis uses dy

Then the side length is a function of \(y\), and the limits are \(y\)-values.

Level 1

Easy

The side length is given directly.

Example

Cross sections perpendicular to the x-axis are squares. The side length is \(s(x)=3-x\) on \([0,3]\). Find the volume.

\(V=\int_0^3 (3-x)^2\,dx\)

Let \(u=3-x\). Then \(u\) runs from 3 to 0.

\(\int_0^3 u^2\,du\)

\(\left[\dfrac{u^3}{3}\right]_0^3\)

\(9\)

Answer: \(9\)

Squares perpendicular to the x-axis have side \(s(x)=x\) on \([0,2]\). Find the volume.

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\(\int_0^2 x^2\,dx\)

\(\dfrac{8}{3}\)

Answer: \(\dfrac{8}{3}\)

The base side in the region is \(\sqrt{x}\) from \(x=0\) to \(x=4\). Cross sections perpendicular to the x-axis are squares. Find the volume.

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Side \(\sqrt{x}\) gives area \(x\).

\(\int_0^4 x\,dx\)

\(8\)

Answer: \(8\)

Isosceles right triangles have leg \(s(x)=x\) on \([0,2]\). The right angle is not in the base. Find the volume.

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Area of one slice: \(\dfrac{1}{2}x^2\).

\(\int_0^2 \dfrac{1}{2}x^2\,dx\)

\(\dfrac{1}{2}\cdot \dfrac{8}{3}\)

\(\dfrac{4}{3}\)

Answer: \(\dfrac{4}{3}\)

Which area formula belongs to a semicircular cross section of radius \(r\)?

A. \(\dfrac{1}{2}\pi r^2\)

B. \(\pi r^2\)

C. \(\dfrac{\sqrt{3}}{4}r^2\)

D. \(2\pi r\)

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A full circle has area \(\pi r^2\).

A semicircle is half of that.

Answer: A. \(\dfrac12\pi r^2\)

Cross sections perpendicular to the y-axis are squares with side \(s(y)=2y\) for \(0\le y\le 3\). Find the volume.

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\(\int_0^3 (2y)^2\,dy\)

\(\int_0^3 4y^2\,dy\)

\(4\cdot 9\)

\(36\)

Answer: \(36\)

Level 2

Medium

The base is a region. Take the side from the two curves.

Example

The base is the region under \(y=\sqrt{x}\) from \(x=0\) to \(x=4\). Cross sections perpendicular to the x-axis are semicircles whose diameter is in the base. Find the volume.

Diameter \(\sqrt{x}\), so radius \(\dfrac{\sqrt{x}}{2}\).

Slice area: \(\dfrac{1}{2}\pi\left(\dfrac{\sqrt{x}}{2}\right)^2\).

\(\dfrac{1}{2}\pi\cdot \dfrac{x}{4}\)

\(\dfrac{\pi x}{8}\)

\(\int_0^4 \dfrac{\pi x}{8}\,dx\)

\(\dfrac{\pi}{8}\cdot 8\)

\(\pi\)

Answer: \(\pi\)

The base is under \(y=2-x\) from \(x=0\) to \(x=2\). Cross sections perpendicular to the x-axis are equilateral triangles with a side in the base. Find the volume.

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Side \(s=2-x\).

\(A(x)=\dfrac{\sqrt{3}}{4}(2-x)^2\)

\(V=\dfrac{\sqrt{3}}{4}\int_0^2 (2-x)^2\,dx\)

The substitution \(u=2-x\) runs from 2 to 0.

\(\dfrac{\sqrt{3}}{4}\int_0^2 u^2\,du\)

\(\dfrac{\sqrt{3}}{4}\cdot \dfrac{8}{3}\)

\(\dfrac{2\sqrt{3}}{3}\)

Answer: \(\dfrac{2\sqrt{3}}{3}\)

The base is the region bounded by \(y=x\) and \(y=x^2\) on \([0,1]\). Squares perpendicular to the x-axis have a side running from the parabola up to the line. Find the volume.

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Side \(s=x-x^2\).

\(V=\int_0^1 (x-x^2)^2\,dx\)

\((x-x^2)^2=x^2-2x^3+x^4\)

\(\left[\dfrac{x^3}{3}-\dfrac{2x^4}{4}+\dfrac{x^5}{5}\right]_0^1\)

\(\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{5}\)

\(\dfrac{10-15+6}{30}\)

\(\dfrac{1}{30}\)

Answer: \(\dfrac{1}{30}\)

Rectangles perpendicular to the x-axis have height three times their base. The base in the region runs from \(y=0\) to \(y=x\) on \([0,2]\). Find the volume.

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Base \(x\), height \(3x\), slice area \(3x^2\).

\(\int_0^2 3x^2\,dx\)

\(8\)

Answer: \(8\)

A student uses radius \(\sqrt{x}\) when the segment in the base is the diameter of a semicircle. What should the radius be?

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The diameter is \(\sqrt{x}\).

The radius is half the diameter.

\(\dfrac{\sqrt{x}}{2}\)

Answer: \(\dfrac{\sqrt{x}}{2}\)

Cross sections perpendicular to the y-axis are squares. The base runs from \(x=y^2\) to \(x=4\) for \(0\le y\le 2\). Find the volume.

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Side \(4-y^2\).

\(\int_0^2 (4-y^2)^2\,dy\)

\((4-y^2)^2=16-8y^2+y^4\)

\(\left[16y-\dfrac{8y^3}{3}+\dfrac{y^5}{5}\right]_0^2\)

\(32-\dfrac{64}{3}+\dfrac{32}{5}\)

\(\dfrac{480-320+96}{15}\)

\(\dfrac{256}{15}\)

Answer: \(\dfrac{256}{15}\)

Level 3

Hard

A diameter, a known region, or slices perpendicular to the y-axis.

Example

The base is a circle \(x^2+y^2=4\). Cross sections perpendicular to the x-axis are squares with a diagonal as the chord of the circle. A diagonal \(d\) gives square area \(d^2/2\). Set up the volume and evaluate it.

At height \(x\), the chord from the lower semicircle to the upper one has length \(2\sqrt{4-x^2}\).

That chord is the diagonal.

\(A(x)=\dfrac{1}{2}\left(2\sqrt{4-x^2}\right)^2\)

\(A(x)=2(4-x^2)\)

\(V=\int_{-2}^{2} 2(4-x^2)\,dx\)

\(4\int_0^2 (4-x^2)\,dx\)

\(4\left[4x-\dfrac{x^3}{3}\right]_0^2\)

\(4\left(8-\dfrac{8}{3}\right)\)

\(\dfrac{64}{3}\)

Answer: \(\dfrac{64}{3}\)

Semicircular cross sections perpendicular to the x-axis have diameter running from \(y=x^2\) to \(y=x\) on \([0,1]\). Find the volume.

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Diameter \(x-x^2\), radius \(\dfrac{x-x^2}{2}\).

\(A(x)=\dfrac{1}{2}\pi\left(\dfrac{x-x^2}{2}\right)^2\)

\(\dfrac{\pi}{8}(x-x^2)^2\)

From the square-side problem, \(\int_0^1 (x-x^2)^2\,dx=\dfrac{1}{30}\).

\(V=\dfrac{\pi}{8}\cdot \dfrac{1}{30}\)

\(\dfrac{\pi}{240}\)

Answer: \(\dfrac{\pi}{240}\)

Equilateral triangles perpendicular to the y-axis have side \(s(y)=4-y\) for \(0\le y\le 4\). Find the volume.

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\(\dfrac{\sqrt{3}}{4}\int_0^4 (4-y)^2\,dy\)

Let \(u=4-y\). The integral of \(u^2\) from 0 to 4 is \(\dfrac{64}{3}\).

\(\dfrac{\sqrt{3}}{4}\cdot \dfrac{64}{3}\)

\(\dfrac{16\sqrt{3}}{3}\)

Answer: \(\dfrac{16\sqrt{3}}{3}\)

The base is bounded by \(y=\sqrt{4-x^2}\) and the x-axis, a semicircle of radius 2. Cross sections perpendicular to the x-axis are squares with side in the base. Find the volume.

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Side \(\sqrt{4-x^2}\).

\(V=\int_{-2}^{2}(4-x^2)\,dx\)

\(2\int_0^2 (4-x^2)\,dx\)

\(2\left(8-\dfrac{8}{3}\right)\)

\(\dfrac{32}{3}\)

Answer: \(\dfrac{32}{3}\)

A solid has cross-sectional area \(A(x)=4x\) for \(1\le x\le 3\). Find the volume.

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\(\int_1^3 4x\,dx\)

\(\left[2x^2\right]_1^3\)

\(18-2\)

\(16\)

Answer: \(16\)

Why is the volume not the area of the base times one sample height when the cross sections change size?

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Each slice has its own area \(A(x)\).

Multiplying the base area by one height double-counts the geometry.

Integrate \(A(x)\) across the direction perpendicular to the slices.

Answer: integrate the changing slice area

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