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

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?

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: 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

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

• This answer adds nothing new, and came later that the more detailed answer above. Feb 10 '19 at 23:44
• @ViktorGlombik I was the first one posting the solution, check out the answer time Feb 10 '19 at 23:49