Volume formed by rotating the region under y = x^2 about the x-axis

Key concept: This is a disk-method problem, but the wording can hide the interval. The curve meets the x-axis at $x=0$, and the vertical line $x=2$ closes the region.

Step by step

We revolve the region bounded by $y=x^2$, $x=2$, and $y=0$ about the x-axis.

1) Describe the region

The curve $y=x^2$ meets $y=0$ at $x=0$, so the interval is

$$0\le x\le 2.$$

2) Use the disk method

Rotating around the x-axis gives disks with radius

$$R(x)=x^2.$$

So the volume is

$$V=\pi\int_{0}^{2}(x^2)^2\,dx$$ $$V=\pi\int_{0}^{2}x^4\,dx$$

3) Integrate

$$V=\pi\left[\frac{x^5}{5}\right]_{0}^{2} =\pi\cdot\frac{32}{5}$$

So the exact volume is

$$\boxed{\frac{32\pi}{5}}$$

The listed choices do not match this value.

Pitfall alert

A frequent trap is to start the integral at $x=2$ because that is the vertical boundary. But the region also needs the point where the curve hits the x-axis, which is $x=0$. Another slip is confusing radius with diameter; the radius here is $x^2$, not $2x^2$.

Try different conditions

If the solid were rotated about the line $y=c$ instead of the x-axis, the radius would become $|x^2-c|$ and the setup would change to washers. If the boundary $x=2$ moved to $x=a$, the answer would become $\pi\int_0^a x^4\,dx = \frac{\pi a^5}{5}$.

Further reading

disk method, x-axis rotation, definite integral