Set up and evaluate the volume
A dome turned about a low line
The teal region lies between the parabola \(y = 5 - x^2\) and the line \(y = 1\). It turns about the dashed line \(y = -1\), which lies below the region.
The distances from \(y = -1\) are \(6 - x^2\) to the parabola and 2 to the line. At \(x = 0\), they are 6 and 2.
The figure fills the solid with washers. A washer is a flat disk with a round hole, and its volume is its area times its thickness. Drag the slider to add washers.
With 4 washers, the total is \(78.25\pi\). Thinner washers fit the solid more closely, so their total approaches the volume, which the integral gives as \(\tfrac{384\pi}{5}\).
About a vertical line
The axis is the vertical dashed line. The washers are perpendicular to it, so they are horizontal, and each one has thickness \(dy\). Both radii are horizontal distances from the line.
Make the washers thinner. Their total approaches the integral of the washer area along the axis.
Notation
At height \(y\), write both radii as distances from the line \(x = h\), as in the earlier skills.
Find the bounds \(c\) and \(d\), the values of \(y\) where the region starts and ends. The washers fill the solid between them.
A washer of thickness \(dy\) has volume \(\pi\big(\cy{R(y)}^2 - \cy{r(y)}^2\big)\,dy\). Square each radius, subtract, and integrate between the bounds, because the integral adds the washers.
\[\cb{V = \int_c^d \pi\big(R(y)^2 - r(y)^2\big)\,dy}\]About a horizontal line \(y = k\), use \(x\), \(dx\) and bounds in \(x\) instead.
If the outer boundary changes at a value \(m\), split the integral at \(m\). Use the outer radius of each piece.
\[\begin{aligned} \cb{V} &\cb{{}= \int_c^m \pi\big(R_1^2 - r_1^2\big)\,dy} \\ &\quad \cb{{}+ \int_m^d \pi\big(R_2^2 - r_2^2\big)\,dy} \end{aligned}\]
When it fails
Swap the two radii. The red labels put \(r\) on the longer segment and \(R\) on the shorter one.
Then \(R^2 - r^2\) is negative at every height, so the integral is negative. A volume cannot be negative, so \(R\) must be the larger distance on the whole interval.
Example 1: an axis above the region
The region lies under \(y = \sqrt{x}\) and above the x-axis, for \(0 \le x \le 4\). It turns about the line \(y = 3\). Find the volume.
The line \(y = 3\) is above the region, so each distance is 3 minus a height. The x-axis is the lower boundary, so it is farther from the line.
\[\cy{R = 3 - 0 = 3}, \qquad \cy{r = 3 - \sqrt{x}}\]Square each radius. Then subtract.
\[\begin{aligned} &3^2 - \left(3 - \sqrt{x}\right)^2 \\ &= 9 - \left(9 - 6\sqrt{x} + x\right) \\ &= 6\sqrt{x} - x \end{aligned}\]The washers stack along the x-axis from \(x = 0\) to \(x = 4\), so those are the bounds.
\[\begin{aligned} \cb{V} &\cb{{}= \pi\int_0^4 \left(6\sqrt{x} - x\right)dx} \\ &\cb{{}= \pi\left[\,4x^{3/2} - \tfrac{x^2}{2}\,\right]_0^4} \\ &\cb{{}= 24\pi} \end{aligned}\]As a check, \(24\pi \approx 75.4\).
Example 2: a vertical axis
The region lies between \(y = \sqrt{x}\) and \(y = \tfrac{x}{2}\). It turns about the line \(x = -1\). Find the volume.
The axis is vertical, so the variable is \(y\). Write each curve as \(x\) in terms of \(y\). The curves meet where \(y^2 = 2y\), at \(y = 0\) and \(y = 2\).
\[x = y^2, \qquad x = 2y\]The region is to the right of the line \(x = -1\), so each distance is an \(x\)-value plus 1. For \(0 < y < 2\), \(2y > y^2\), so the line \(x = 2y\) is farther.
\[\cy{R = 2y + 1}, \qquad \cy{r = y^2 + 1}\]Square each radius. Then subtract.
\[\begin{aligned} &(2y + 1)^2 - (y^2 + 1)^2 \\ &= 2y^2 + 4y - y^4 \end{aligned}\]A tempting wrong move: The region runs from \(x = 0\) to \(x = 4\), so integrate from 0 to 4.
\[\begin{aligned} &\pi\int_0^4 \left(2y^2 + 4y - y^4\right)dy \\ &= -\tfrac{1952\pi}{15} \end{aligned}\]The volume is negative. The integral is in \(y\), so its bounds are \(y\)-values, and in the picture the washers stack from \(y = 0\) to \(y = 2\).
Use the \(y\)-values where the curves meet, 0 and 2.
\[\begin{aligned} \cb{V} &\cb{{}= \pi\int_0^2 \left(2y^2 + 4y - y^4\right)dy} \\ &\cb{{}= \pi\left[\,\tfrac{2y^3}{3} + 2y^2 - \tfrac{y^5}{5}\,\right]_0^2} \\ &\cb{{}= \tfrac{104\pi}{15}} \end{aligned}\]
Practice
The region lies between \(y = x^3\) and \(y = x\), for \(0 \le x \le 1\). It turns about the line \(y = -2\). Find the volume.
Slide the slice to \(x = \tfrac12\). Is the axis above or below the region? Which boundary is farther from it?
Solution
The axis is below the region, so each distance is a height plus 2. On \((0, 1)\), the line \(y = x\) is above the curve \(y = x^3\), so the line is farther.
\[\cy{R = x + 2}, \qquad \cy{r = x^3 + 2}\]Square each radius. Then subtract.
\[\begin{aligned} &(x + 2)^2 - (x^3 + 2)^2 \\ &= x^2 + 4x - 4x^3 - x^6 \end{aligned}\]Integrate from 0 to 1.
\[\begin{aligned} \cb{V} &\cb{{}= \pi\left[\,\tfrac{x^3}{3} + 2x^2 - x^4 - \tfrac{x^7}{7}\,\right]_0^1} \\ &\cb{{}= \tfrac{25\pi}{21}} \end{aligned}\]
Questions
The graph shows the region between \(y = 2x\) and \(y = 6\), for \(0 \le x \le 3\). It turns about the line \(y = 8\).
Which integral gives the volume of the solid?
- \(\pi\int_0^3 \left((8 - 2x)^2 - 2^2\right)dx\)
- \(\pi\int_0^3 \left((2x)^2 - 6^2\right)dx\)
- \(\pi\int_0^3 \left(8 - 2x - 2\right)^2 dx\)
- \(\pi\int_0^6 \left((8 - 2x)^2 - 2^2\right)dx\)
Answer
\(\pi\int_0^3 \left((8 - 2x)^2 - 2^2\right)dx\)
What is the inner radius \(r(x)\)?
- 2, at every \(x\)
- 6
- \(8 - 2x\)
Answer
2, at every \(x\)
The region between \(y = 1\) and \(y = 1 + x\), for \(0 \le x \le 2\), turns about the line \(y = -1\). What is the volume?
- \(\dfrac{32\pi}{3}\)
- \(\dfrac{8\pi}{3}\)
- \(\dfrac{20\pi}{3}\)
Answer
\(\dfrac{32\pi}{3}\)
Summary
You write both radii as distances from the shifted axis, square each one, subtract, and integrate over the bounds along the axis.
This works because each washer is the outer disk without the hole, and the volume is the total of the washers.
You use it for any region that turns about a line, and you leave the answer as a multiple of \(\pi\).
This is one of four skills in topic 8.12 of Which & Why: AP Calculus.