Find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $x=5$, and $y=0$ about the $x$-axis

Question

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $x=5$, and $y=0$ about the $x$-axis.

How do I solve this when given rotating region about the x-axis?

Answer 1

Your bounds of integration are going to be from $0$ to $5$. Your radius function is going to be $r(x)=x^2$. Now, just plug it into the formula for the disk method which is $V=\pi\int_{a}^{b}\left[r(x)\right]^2\,dx$ and you're good to go:

$$ V=\pi\int_{0}^{5}(x^2)^2\,dx= \pi\int_{0}^{5}x^4\,dx=\pi\frac{x^5}{5}\bigg|_{0}^{5}=\pi\left(\frac{5^5}{5}-\frac{0}{5}\right)=625\pi. $$

Here is a rough picture of what you are doing:

Answer 2

Using the formula $V = \pi \int_a^b f^2(x) dx$ , the volume you are looking for is the volume gerenated by the curve between $x=0$ and $x=5$, so the volume is

$V = \pi \int_0^5 x^4 dx = 625 \pi$