Set up and evaluate the volume
Adding up the washers
The teal region lies between the parabola \(y = 4x - x^2\) and the line \(y = x\), from \(x = 0\) to \(x = 3\). It turns about the x-axis.
The figure fills the solid with a stack of washers. A washer is a thin slab, so its volume is its area times its thickness.
Move the slider to add washers. The readout adds the volumes of all the washers. With 3 washers, the total is about \(22.44\pi\). With 40 washers, it is about \(21.60\pi\).
Thinner washers fit the solid more closely, so the total comes closer to the volume of the solid.
Any washer solid
This region has no numbers. A gap separates it from the axis, so each slice sweeps a washer.
A washer at \(x\) has an area \(A(x)\) and a thickness \(\Delta x\), so its volume is \(A(x)\,\Delta x\).
Make the washers thinner. Their total comes closer to one number, the volume of the solid. That number is the integral of the washer area from one end of the region to the other.
Notation
A washer at \(x\) has outer radius \(\cy{R(x)}\), inner radius \(\cy{r(x)}\) and thickness \(dx\). Its volume is its area times its thickness.
\[\cb{\pi\big(R(x)^2 - r(x)^2\big)}\,dx\]The integral adds the volumes of all the washers from \(\cy{a}\) to \(\cy{b}\).
\[\cb{V = \int_{\cy{a}}^{\cy{b}} \pi\big(R(x)^2 - r(x)^2\big)\,dx}\]To evaluate the integral, square each radius and subtract. Then integrate each term, and keep \(\pi\) in the answer.
About a vertical axis, the variable is \(y\), and the integral ends in \(dy\).
When it fails
The red disk has radius \(R - r\), so its area is \(\pi(R - r)^2\). Integrating this area from 0 to 3 gives \(\tfrac{81\pi}{10}\).
The solid has volume \(\tfrac{108\pi}{5}\), which is larger. The washer area is the outer disk minus the hole, not a disk of radius \(R - r\).
Example 1: a horizontal axis
The region between \(y = \sqrt{x}\) and \(y = 3\), to the right of the y-axis, turns about the x-axis. Find the volume.
The curves meet where \(\sqrt{x} = 3\), at \(x = 9\). The y-axis closes the region on the left, so the bounds are 0 and 9.
\[0 \le x \le 9\]The line is farther from the x-axis than the curve, so \(R = 3\) and \(r = \sqrt{x}\). Square each radius and subtract.
\[\begin{aligned} \cb{V} &= \pi\int_0^9 \Big(3^2 - \big(\sqrt{x}\big)^2\Big)\,dx \\ &= \pi\int_0^9 (9 - x)\,dx \end{aligned}\]Integrate, and substitute the bounds.
\[\begin{aligned} \cb{V} &= \pi\Big[\,9x - \tfrac{x^2}{2}\,\Big]_0^9 \\ &= \pi\Big(81 - \tfrac{81}{2}\Big) = \tfrac{81\pi}{2} \end{aligned}\]As a check, \(\tfrac{81\pi}{2}\) is about 127.2.
Example 2: a vertical axis
The region between \(x = y^2 + 2\) and \(x = 6\) turns about the y-axis. Find the volume.
The axis is vertical, so the variable is \(y\). The curves meet where \(y^2 + 2 = 6\), which gives \(y = -2\) and \(y = 2\).
The line \(x = 6\) is farther from the axis than the parabola, so \(R = 6\) and \(r = y^2 + 2\).
A tempting wrong move: The solid is the region turned about the axis, so use only the outer radius.
\[V = \pi\int_{-2}^{2} 6^2\,dy = 144\pi\]This counts the hole, shown in red, as part of the solid. The hole holds no material, so the true volume is less than \(144\pi\).
Subtract the area of the hole. Expand its square before you integrate.
\[\begin{aligned} \cb{V} &= \pi\int_{-2}^{2} \big(36 - (y^2 + 2)^2\big)\,dy \\ &= \pi\int_{-2}^{2} \big(32 - 4y^2 - y^4\big)\,dy \end{aligned}\]Integrate, and substitute the bounds.
\[\begin{aligned} \cb{V} &= \pi\Big[\,32y - \tfrac{4y^3}{3} - \tfrac{y^5}{5}\,\Big]_{-2}^{2} \\ &= \tfrac{1408\pi}{15} \end{aligned}\]As a check, \(\tfrac{1408\pi}{15}\) is about \(93.9\pi\), which is less than \(144\pi\).
Practice
The region between \(y = x^2\) and \(y = 6 - x\) turns about the x-axis. Find the volume.
Drag the slice to \(x = 1\). Which curve is farther from the x-axis? That curve gives \(R\).
Solution
The curves meet where \(x^2 = 6 - x\), so the bounds are \(-3\) and \(2\).
\[x^2 + x - 6 = (x + 3)(x - 2) = 0\]The line is farther from the axis, so \(R = 6 - x\) and \(r = x^2\). Square each radius and subtract.
\[\begin{aligned} \cb{V} &= \pi\int_{-3}^{2} \big((6 - x)^2 - x^4\big)\,dx \\ &= \pi\int_{-3}^{2} \big(36 - 12x \\ &\qquad\qquad + x^2 - x^4\big)\,dx \end{aligned}\]Integrate, and substitute the bounds.
\[\begin{aligned} \cb{V} &= \pi\Big[\,36x - 6x^2 + \tfrac{x^3}{3} - \tfrac{x^5}{5}\,\Big]_{-3}^{2} \\ &= \tfrac{500\pi}{3} \end{aligned}\]
Questions
The region between \(y = x + 2\) and \(y = 6\), for \(0 \le x \le 4\), turns about the x-axis.
Which integral gives the volume of the solid?
- \(\pi\displaystyle\int_0^4 \big(36 - (x + 2)^2\big)\,dx\)
- \(\pi\displaystyle\int_0^4 \big(6 - (x + 2)\big)^2\,dx\)
- \(\pi\displaystyle\int_0^4 36\,dx\)
- \(\pi\displaystyle\int_2^6 \big(36 - (x + 2)^2\big)\,dx\)
Answer
\(\pi\displaystyle\int_0^4 \big(36 - (x + 2)^2\big)\,dx\)
In the integrand, what does the term \(\pi(x + 2)^2\) measure?
- The area of the hole in the slice at \(x\)
- The area of the washer
- The thickness of the slice
Answer
The area of the hole in the slice at \(x\)
The region between \(y = 3x\) and \(y = x^2\) turns about the x-axis. What is the volume?
- \(\dfrac{162\pi}{5}\)
- \(\dfrac{81\pi}{10}\)
- \(81\pi\)
Answer
\(\dfrac{162\pi}{5}\)
Summary
You write the washer area in the slicing variable, and you integrate it between the bounds.
This works because the volume is the total of the washer volumes, and each volume is its area times its thickness.
You use it for a solid with a hole that turns about a coordinate axis, and you keep \(\pi\) in the answer.
This is one of four skills in topic 8.11 of Which & Why: AP Calculus.