detineerlf

2022-07-16

AP Calc AB Problem - Finding volume
The base of a solid in the region bounded by the two parabolas ${y}^{2}=8x$ and ${x}^{2}=8y$. Cross sections of the solid perpendicular to the x-axis are semicircles. What is the volume, in cubic units, of the solid?
The answer choices are:
A) $\frac{288\pi }{35}$
B) $\frac{576\pi }{35}$
C) $\frac{144\pi }{35}$
D) $8\pi$
I started off by writing the integral like this:
${\int }_{0}^{8}\frac{1}{2}\pi \left(\frac{1}{2}\left(\sqrt{8x}-\frac{{x}^{2}}{8}\right){\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$
then I simplified it to $\frac{\pi }{8}{\int }_{0}^{8}\left(\sqrt{8x}-\frac{{x}^{2}}{8}{\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$
I can't think of anyway to solve this problem from here, without using a calculator.

Expert

Step 1
Expand the binomial term:
$V=\frac{\pi }{8}{\int }_{x=0}^{8}{\left(\sqrt{8x}-\frac{{x}^{2}}{8}\right)}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi }{8}{\int }_{x=0}^{8}8x-2\sqrt{8x}\cdot \frac{{x}^{2}}{8}+\frac{{x}^{4}}{64}\phantom{\rule{thinmathspace}{0ex}}dx.$
Step 2
Now integrate term by term:
$V=\frac{\pi }{8}{\left[4{x}^{2}-\frac{\sqrt{2}}{7}{x}^{7/2}+\frac{{x}^{5}}{5\left(64\right)}\right]}_{x=0}^{8}.$

Do you have a similar question?

Recalculate according to your conditions!