Set up the disk integral
Three parts fill the integral
The teal region lies between the parabola \(y = x^2\) and the line \(y = 1\), for \(-1 \le x \le 1\). It turns about the dashed line, and the slider moves a slice.
At \(x = 0.5\), the parabola is at height 0.25 and the axis is at height 1, so the radius is the gap, 0.75.
The disk area is \(\pi\) times the radius squared, so at \(x = 0.5\) it is \(\frac{9\pi}{16}\), about \(0.56\pi\).
At every \(x\) the radius is \(1 - x^2\), so the disk area is \(\pi(1 - x^2)^2\). The integral adds these areas from \(x = -1\) to \(x = 1\), the two ends of the region. Press "Stack the disks" to fill the solid with thin disks.
The variable, the squared radius and the bounds
The integral has three parts, and each part comes from one mark on the picture.
The variable is the coordinate that runs along the axis, and the integral ends with it, as \(dx\).
The integrand is the expression after the integral sign. It is the area of the disk, \(\pi\) times the squared orange radius, written in the same variable.
The bounds are the two ends of the region along the axis, \(a\) and \(b\). The slider moves the disk from one end to the other.
Notation
Choose the variable. It runs along the axis, so it is \(x\) for a horizontal axis and \(y\) for a vertical axis.
Write the radius \(R(x)\) in that variable. The integrand squares the whole radius, not the curve alone.
\[\begin{aligned} \cy{R(x)} = {}&\text{the distance from} \\ &\text{the axis to the curve} \end{aligned}\]The integrand is the area of one disk, and the bounds are the first and last value of the variable on the region.
\[V = \int_{\cy{a}}^{\cy{b}} \pi\,\cy{R(x)}^2\,dx\]A bound is often a point where the curve meets the axis. There the radius is 0, so set the radius equal to 0 and solve.
\[\cy{R(x)} = 0\]
When it fails
The axis must touch the region. Press the button to move the region off the line.
A gap opens between the axis and the region, so each slice is a ring with an outer and an inner orange radius. The second button colors its empty center red. The integrand \(\pi R^2\) counts that red center as solid, so this lesson covers only regions that touch the axis.
Example 1: a horizontal axis
The region lies between the curve \(y = \dfrac{4}{x}\), the line \(y = 4\) and the line \(x = 4\). It turns about the line \(y = 4\). Write the integral for the volume.
The axis \(y = 4\) is horizontal, so the slices are vertical and the variable is \(x\).
The curve is below the axis, so the radius is the axis height minus the curve height.
\[\cy{R(x)} = 4 - \dfrac{4}{x}\]The curve meets the axis where \(\frac{4}{x} = 4\), at \(x = 1\). The line \(x = 4\) closes the region on the right, so the bounds are 1 and 4.
\[1 \le x \le 4\]Square the radius, multiply by \(\pi\), and integrate between the bounds.
\[V = \int_1^4 \pi\left(4 - \dfrac{4}{x}\right)^2 dx\]
Example 2: a vertical axis
The region lies between the curve \(x = 1 + y^2\) and the line \(x = 5\). It turns about the line \(x = 5\). Write the integral for the volume.
The axis \(x = 5\) is vertical, so the slices are horizontal and the variable is \(y\).
A tempting wrong move: Square the x-value of the curve.
\[V = \int_{-2}^{2} \pi\left(1 + y^2\right)^2 dy\]At \(y = 0\) the curve is at \(x = 1\) and the axis is at \(x = 5\), so the orange radius is 4. The red disk uses the curve's x-value, 1, and is too small.
The curve is to the left of the axis, so subtract the curve position from the axis position.
\[\cy{R(y)} = 5 - \left(1 + y^2\right) = 4 - y^2\]The curve meets the axis where \(1 + y^2 = 5\), at \(y = \pm 2\), so these are the bounds.
\[-2 \le y \le 2\]Square the radius, multiply by \(\pi\), and integrate between the bounds.
\[V = \int_{-2}^{2} \pi\left(4 - y^2\right)^2 dy\]
Practice
The region lies between the curve \(y = 1 + x^2\), the line \(y = 5\) and the y-axis. It turns about the line \(y = 5\). Write the integral for the volume.
Move the slider. The orange segment runs from the axis down to the curve.
Solution
The axis is horizontal, so the variable is \(x\). The curve is below the axis, so the radius is the axis height minus the curve height.
\[\cy{R(x)} = 5 - \left(1 + x^2\right) = 4 - x^2\]The y-axis, \(x = 0\), is the left end. The curve meets the axis where \(1 + x^2 = 5\), at \(x = 2\), so the bounds are 0 and 2.
\[0 \le x \le 2\]Square the radius and multiply by \(\pi\).
\[V = \int_0^2 \pi\left(4 - x^2\right)^2 dx\]
Questions
The region between the curve \(x = \dfrac{y^2}{4}\) and the line \(x = 4\), for \(-4 \le y \le 4\), turns about the line \(x = 4\).
Which variable and which bounds does the integral use?
- \(y\), from \(-4\) to 4
- \(x\), from 0 to 4
- \(y\), from 0 to 4
Answer
\(y\), from \(-4\) to 4
What is the radius of the disk at \(y = 2\)?
- 3
- 1
- 4
- 9
Answer
3
The region lies between the curve \(y = 2x^2\), the line \(y = 18\) and the y-axis. It turns about the line \(y = 18\). Which integral gives the volume?
- \(\displaystyle\int_0^3 \pi(18 - 2x^2)^2\,dx\)
- \(\displaystyle\int_0^3 \pi(2x^2)^2\,dx\)
- \(\displaystyle\int_0^{18} \pi(18 - 2x^2)^2\,dx\)
Answer
\(\displaystyle\int_0^3 \pi(18 - 2x^2)^2\,dx\)
Summary
You write the variable along the axis, the squared radius measured from the axis, and the bounds where the region begins and ends.
This works because each disk has an area of \(\pi\) times its squared distance from the axis, and the integral adds the disks.
Use it to turn a region and an axis into an integral before you evaluate anything.
This is one of four skills in topic 8.10 of Which & Why: AP Calculus.