AP Calculus AB · Lesson 3

Volumes by cross sections and revolution

Know Know it

Slice a solid perpendicular to the x-axis. If each slice at x has area A(x)A(x), the volume is ∫abA(x) dx\int_a^b A(x)\,dx. For a base region between two curves, the side or diameter of each slice is top minus bottom.

Revolve a region about a horizontal line and each slice is a disc or a washer. Disc: V=π∫abR(x)2 dxV = \pi\int_a^b R(x)^2\,dx. Washer: V=π∫ab(R(x)2−r(x)2)dxV = \pi\int_a^b \left(R(x)^2 - r(x)^2\right)dx, with R the outer radius and r the inner radius.

A radius is a distance from the axis. About the line y=ky = k, a curve at height y=f(x)y = f(x) gives the radius ∣f(x)−k∣|f(x) - k|. About y=−1y = -1, a curve at height y=x2y = x^2 gives x2+1x^2 + 1.

Area of a slice whose side or diameter is s
Shape of the sliceArea
Square with side ss2s^2
Semicircle with diameter sπs28\frac{\pi s^2}{8}
Equilateral triangle with side s34s2\frac{\sqrt{3}}{4}s^2
Isosceles right triangle with a leg ss22\frac{s^2}{2}

Trap Square each radius, then subtract

A washer's area is πR2−πr2\pi R^2 - \pi r^2, not π(R−r)2\pi(R - r)^2. With R=3R = 3 and r=2r = 2, the washer has area 9π−4π=5π9\pi - 4\pi = 5\pi, while π(3−2)2=π\pi(3 - 2)^2 = \pi is far too small.

Worked example Easy

The region under y=x+1y = x + 1 for 0≤x≤20 \le x \le 2 is revolved about the x-axis. What is the volume of the solid?

  1. 4π4\pi
  2. 8π3\frac{8\pi}{3}
  3. 26π3\frac{26\pi}{3}Answer
  4. 263\frac{26}{3}

How to solve it

  1. A slice at x is a disc with radius R(x)=x+1R(x) = x + 1, so its area is π(x+1)2\pi(x + 1)^2.
  2. V=π∫02(x+1)2 dx=π[(x+1)33]02=π(9−13)=26π3V = \pi\int_0^2 (x + 1)^2\,dx = \pi\left[\frac{(x + 1)^3}{3}\right]_0^2 = \pi\left(9 - \frac{1}{3}\right) = \frac{26\pi}{3}.

Why each choice is right or wrong

  • A. Incorrect. 4π4\pi is π∫02(x+1) dx\pi\int_0^2 (x + 1)\,dx, which forgets to square the radius.
  • B. Incorrect. 8π3\frac{8\pi}{3} is π∫02x2 dx\pi\int_0^2 x^2\,dx, which drops the +1+1 from the radius.
  • C. Correct. Each slice is a disc of radius x+1x + 1, so V=π∫02(x+1)2 dx=π⋅27−13=26π3V = \pi\int_0^2 (x + 1)^2\,dx = \pi \cdot \frac{27 - 1}{3} = \frac{26\pi}{3}.
  • D. Incorrect. 263\frac{26}{3} is ∫02(x+1)2 dx\int_0^2 (x + 1)^2\,dx without the factor π\pi from the area of each disc.

Worked example Medium

The base of a solid is the region between y=4−x2y = 4 - x^2 and the x-axis. Each cross section perpendicular to the x-axis is a square with one side in the base. What is the volume of the solid?

  1. 323\frac{32}{3}
  2. 25615\frac{256}{15}
  3. 512π15\frac{512\pi}{15}
  4. 51215\frac{512}{15}Answer

How to solve it

  1. The base runs from x=−2x = -2 to x=2x = 2, where 4−x2=04 - x^2 = 0.
  2. The side of the square at x is s=4−x2s = 4 - x^2, so A(x)=(4−x2)2=16−8x2+x4A(x) = (4 - x^2)^2 = 16 - 8x^2 + x^4.
  3. V=∫−22(16−8x2+x4) dx=64−1283+645=51215V = \int_{-2}^{2} (16 - 8x^2 + x^4)\,dx = 64 - \frac{128}{3} + \frac{64}{5} = \frac{512}{15}.

Why each choice is right or wrong

  • A. Incorrect. 323\frac{32}{3} is ∫−22(4−x2) dx\int_{-2}^{2} (4 - x^2)\,dx, the area of the base, which forgets to square the side.
  • B. Incorrect. 25615\frac{256}{15} integrates only from 0 to 2, which is half of the base.
  • C. Incorrect. 512π15\frac{512\pi}{15} puts in a factor π\pi, but the slices are squares, not discs.
  • D. Correct. Each square has side 4−x24 - x^2, so V=∫−22(4−x2)2 dx=51215V = \int_{-2}^{2} (4 - x^2)^2\,dx = \frac{512}{15}.

Explain Explain it

Task verbs: construct, calculate

Build the integral in three moves: sketch one slice, write its area as a function of x, then choose the limits from the base region. Name the outer and inner radius before you square them.

For a washer about y=ky = k with the region above the axis, the outer radius uses the curve farther from the axis: R=(far curve)−kR = (\text{far curve}) - k and r=(near curve)−kr = (\text{near curve}) - k.

Worked example Hard

The region bounded by y=xy = x and y=x2y = x^2 is revolved about the line y=−1y = -1. What is the volume of the solid?

  1. 7π15\frac{7\pi}{15}Answer
  2. π30\frac{\pi}{30}
  3. 2π15\frac{2\pi}{15}
  4. π6\frac{\pi}{6}

How to solve it

  1. The curves meet at x=0x = 0 and x=1x = 1, and y=xy = x is on top there.
  2. Distances from y=−1y = -1: outer radius R=x−(−1)=x+1R = x - (-1) = x + 1, inner radius r=x2+1r = x^2 + 1.
  3. V=π∫01((x+1)2−(x2+1)2)dx=π∫01(2x−x2−x4) dx=π(1−13−15)=7π15V = \pi\int_0^1 \left((x + 1)^2 - (x^2 + 1)^2\right)dx = \pi\int_0^1 (2x - x^2 - x^4)\,dx = \pi\left(1 - \frac{1}{3} - \frac{1}{5}\right) = \frac{7\pi}{15}.

Why each choice is right or wrong

  • A. Correct. The outer radius is x+1x + 1 and the inner radius is x2+1x^2 + 1, so V=π∫01((x+1)2−(x2+1)2)dx=7π15V = \pi\int_0^1 \left((x + 1)^2 - (x^2 + 1)^2\right)dx = \frac{7\pi}{15}.
  • B. Incorrect. π30\frac{\pi}{30} is π∫01(x−x2)2 dx\pi\int_0^1 (x - x^2)^2\,dx, which squares the difference of the radii instead of subtracting the squares.
  • C. Incorrect. 2π15\frac{2\pi}{15} is π∫01(x2−x4) dx\pi\int_0^1 (x^2 - x^4)\,dx, the volume about the x-axis; it measures the radii from y=0y = 0, not from y=−1y = -1.
  • D. Incorrect. π6\frac{\pi}{6} is π∫01(x−x2) dx\pi\int_0^1 (x - x^2)\,dx, which never squares the radii.

Construct Your turn

The base of a solid is the region in the first quadrant bounded by the x-axis, the y-axis and the line x+y=3x + y = 3. Each cross section perpendicular to the x-axis is a semicircle whose diameter lies in the base. Write an integral for the volume of the solid, then evaluate it.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the area of a cross section as π8(3−x)2\frac{\pi}{8}(3 - x)^2
  • Integrates that area from x=0x = 0 to x=3x = 3
  • Evaluates the volume as 9π8\frac{9\pi}{8}

Model answer

At x, the base runs from the x-axis up to the line y=3−xy = 3 - x, so the diameter of the semicircle is 3−x3 - x and its radius is 3−x2\frac{3 - x}{2}. Its area is 12π(3−x2)2=π8(3−x)2\frac{1}{2}\pi\left(\frac{3 - x}{2}\right)^2 = \frac{\pi}{8}(3 - x)^2. The volume is ∫03π8(3−x)2 dx=π8[−(3−x)33]03=π8⋅9=9π8\int_0^3 \frac{\pi}{8}(3 - x)^2\,dx = \frac{\pi}{8}\left[-\frac{(3 - x)^3}{3}\right]_0^3 = \frac{\pi}{8} \cdot 9 = \frac{9\pi}{8}.

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