Build the disk area
Double the radius, quadruple the area
The teal region lies under the line \(y = \tfrac{x}{2}\) from \(x = 0\) to \(x = 6\). It turns about the x-axis and forms a cone. The slider moves a slice through the cone.
The slice is a disk, and the orange segment is its radius. The area of a disk is \(\pi\) times the square of the radius.
At \(x = 2\) the radius is 1, so the area is \(\pi\). At \(x = 4\) the radius is 2, so the area is \(4\pi\).
The radius doubles and the area becomes four times as large, because the area depends on the square of the radius.
The area is pi times the radius squared
The slice is a disk with the orange radius \(\cy{r}\), and its teal face has the area \(\pi r^2\).
The radius changes from slice to slice, so the area changes too. About the x-axis you write the radius as \(r(x)\) and the area as \(A(x)\).
Press "Turn about the y-axis". The same rule holds, with the radius \(r(y)\) and the area \(A(y)\).
The disk is solid from the axis to its edge, so there is nothing to subtract. Whatever the slice, you square the whole radius and multiply by \(\pi\).
Notation
The slice at \(x\) is a disk. Its radius is the distance \(\cy{r(x)}\) from the axis to the curve.
The area of a disk is pi times the radius squared.
\[\cb{A(x)} = \pi\,[\cy{r(x)}]^2\]Square the whole radius. If the radius is a sum of two terms, its square has three terms.
\[(a + b)^2 = a^2 + 2ab + b^2\]For slices perpendicular to the y-axis, the same rule holds with y.
\[\cb{A(y)} = \pi\,[\cy{r(y)}]^2\]
When it fails
A solid disk needs the region to reach the axis. Press the button to cut a hole: the region lifts off the x-axis.
Each slice is now a ring, and its red center is empty. The formula \(\pi r^2\) counts the red center as solid, so it gives too much area. This lesson covers regions that reach the axis, so each slice is a solid disk.
Example 1: square a power
The region lies under \(y = x^{3/2}\) from \(x = 0\) to \(x = 4\). It turns about the x-axis. Write the area of the disk at \(x\).
The axis is the x-axis, so the radius is the height of the curve.
\[\cy{r(x)} = x^{3/2}\]Square the radius. To square a power, double its exponent.
\[\left(x^{3/2}\right)^2 = x^3\]Multiply by \(\pi\).
\[\cb{A(x)} = \pi x^3\]At \(x = 2\) the radius squared is 8, so the area is \(8\pi\).
\[\pi \cdot 2^3 = 8\pi\]
Example 2: square a sum
The region lies under \(y = 1 + \sqrt{x}\) from \(x = 0\) to \(x = 4\). It turns about the x-axis. Write the area of the disk at \(x\).
The axis is the x-axis, so the radius is the height of the curve.
\[\cy{r(x)} = 1 + \sqrt{x}\]A tempting wrong move: Square each term of the radius. The area is \(\pi(1 + x)\).
\[\begin{aligned} A(x) &= \pi\left(1^2 + (\sqrt{x})^2\right) \\ &= \pi(1 + x) \end{aligned}\]At \(x = 4\) the orange radius is 3, so the disk has area \(9\pi\). The formula gives \(5\pi\), the area of the red disk, which is too small. The middle term \(2\sqrt{x}\) is missing.
Square the whole radius. The square of a sum has a middle term.
\[(1 + \sqrt{x})^2 = 1 + 2\sqrt{x} + x\]Multiply by \(\pi\). At \(x = 4\) the area is \(9\pi\), the area of a disk of radius 3.
\[\begin{aligned} \cb{A(x)} &= \pi\left(1 + 2\sqrt{x} + x\right) \\ \cb{A(4)} &= \pi(1 + 4 + 4) = 9\pi \end{aligned}\]
Practice
The region lies under \(y = 3 - x^2\) from \(x = 0\) to \(x = \sqrt{3}\). It turns about the x-axis. Write the area of the disk at \(x\).
Move the slider. The radius is the height of the curve, \(3 - x^2\). Square all of it. Then multiply by \(\pi\).
Solution
The radius is the height of the curve.
\[\cy{r(x)} = 3 - x^2\]Square the whole radius, with its middle term.
\[(3 - x^2)^2 = 9 - 6x^2 + x^4\]Multiply by \(\pi\). At \(x = 1\) the radius is 2, so the area is \(4\pi\).
\[\begin{aligned} \cb{A(x)} &= \pi\left(9 - 6x^2 + x^4\right) \\ \cb{A(1)} &= \pi(9 - 6 + 1) = 4\pi \end{aligned}\]
Questions
The figure shows the region under \(y = \sqrt{25 - x^2}\) from \(x = 0\) to \(x = 5\). It turns about the x-axis.
What is the area of the disk at \(x = 3\)?
- \(16\pi\)
- \(4\pi\)
- \(8\pi\)
- 16
Answer
\(16\pi\)
At which value of \(x\) is the area of the disk \(9\pi\)?
- 4
- 3
- 9
- 5
Answer
4
The region under \(y = x + 2\) from \(x = 0\) to \(x = 3\) turns about the x-axis. What is the area of the disk at \(x = 1\)?
- \(9\pi\)
- \(5\pi\)
- \(3\pi\)
- \(6\pi\)
Answer
\(9\pi\)
Summary
You square the whole radius and multiply by pi to find the area of a disk.
This works because a disk is solid from the axis to its edge, so its area has no part to subtract.
Use it for every slice of a region that reaches the axis, with the radius measured at that slice.
This is one of four skills in topic 8.9 of Which & Why: AP Calculus.