How to find volume when a region is rotated around the x-axis

Key concept: When a region sits between two curves and spins around the x-axis, washers are usually the right tool. The only real work is identifying the outer and inner radii correctly.

Step by step

Step 1: Choose the washer method

Region $R$ is between the curves from $x=0$ to $x=2$, and it is rotated about the x-axis.

So the volume is

$V=\pi\int_0^2\left(R(x)^2-r(x)^2\right)dx$

where

• outer radius: $R(x)=8-2x$
• inner radius: $r(x)=\frac12 x^2$

Step 2: Set up the integral

$V=\pi\int_0^2\left((8-2x)^2-\left(\frac12 x^2\right)^2\right)dx$

Step 3: Expand

$(8-2x)^2=64-32x+4x^2$

$\left(\frac12 x^2\right)^2=\frac14 x^4$

So

$V=\pi\int_0^2\left(64-32x+4x^2-\frac14 x^4\right)dx$

Step 4: Integrate

$V=\pi\left[64x-16x^2+\frac43 x^3-\frac{x^5}{20}\right]_0^2$

$V=\pi\left(128-64+\frac{32}{3}-\frac{32}{20}\right)$

$V=\pi\left(64+\frac{32}{3}-\frac85\right)=\pi\cdot\frac{1096}{15}$

Answer

$\boxed{\frac{1096\pi}{15}}$

Pitfall alert

A frequent slip is using the region height as the radius. For rotation about the x-axis, radii are y-values, so you need squares of the functions, not just their difference.

Try different conditions

If the same region were rotated about the y-axis instead, you would likely switch to cylindrical shells or rewrite the curves in terms of $y$. The setup would look completely different.

Further reading

washer method, volume of revolution, definite integral