detineerlf

Answered

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:

${\int}_{0}^{8}\frac{1}{2}\pi {\textstyle (}\frac{1}{2}{\textstyle (}\sqrt{8x}-\frac{{x}^{2}}{8}{\textstyle )}{{\textstyle )}}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$

then I simplified it to $\frac{\pi}{8}{\int}_{0}^{8}{\textstyle (}\sqrt{8x}-\frac{{x}^{2}}{8}{{\textstyle )}}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$

I can't think of anyway to solve this problem from here, without using a calculator.

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:

${\int}_{0}^{8}\frac{1}{2}\pi {\textstyle (}\frac{1}{2}{\textstyle (}\sqrt{8x}-\frac{{x}^{2}}{8}{\textstyle )}{{\textstyle )}}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$

then I simplified it to $\frac{\pi}{8}{\int}_{0}^{8}{\textstyle (}\sqrt{8x}-\frac{{x}^{2}}{8}{{\textstyle )}}^{2}\phantom{\rule{thinmathspace}{0ex}}dx$

I can't think of anyway to solve this problem from here, without using a calculator.

Answer & Explanation

slapadabassyc

Expert

2022-07-17Added 21 answers

Step 1

Expand the binomial term:

$V=\frac{\pi}{8}{\int}_{x=0}^{8}{(\sqrt{8x}-\frac{{x}^{2}}{8})}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi}{8}{\int}_{x=0}^{8}8x-2\sqrt{8x}\cdot \frac{{x}^{2}}{8}+\frac{{x}^{4}}{64}\phantom{\rule{thinmathspace}{0ex}}dx.$

Step 2

Now integrate term by term:

$V=\frac{\pi}{8}{[4{x}^{2}-\frac{\sqrt{2}}{7}{x}^{7/2}+\frac{{x}^{5}}{5(64)}]}_{x=0}^{8}.$

Expand the binomial term:

$V=\frac{\pi}{8}{\int}_{x=0}^{8}{(\sqrt{8x}-\frac{{x}^{2}}{8})}^{2}\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi}{8}{\int}_{x=0}^{8}8x-2\sqrt{8x}\cdot \frac{{x}^{2}}{8}+\frac{{x}^{4}}{64}\phantom{\rule{thinmathspace}{0ex}}dx.$

Step 2

Now integrate term by term:

$V=\frac{\pi}{8}{[4{x}^{2}-\frac{\sqrt{2}}{7}{x}^{7/2}+\frac{{x}^{5}}{5(64)}]}_{x=0}^{8}.$

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