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November 4th, 2017, 06:22 AM   #1
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surface integral problem

Find the area of the cylinder $\displaystyle x^2 + z^2 = a^2$ that lies inside the cylinder $\displaystyle x^2 + y^2 = a^2$.

Official Answer: $\displaystyle 8a^2$

My solution
$\displaystyle
x = r cos \theta , y = r sin \theta , z = \sqrt{a^2 - r^2cos^2 \theta }
$

$\displaystyle
\int_{0}^{2 \pi} \int_{0}^{a} \sqrt { \frac {(r^2cos \theta sin^2 \theta + r^2 cos^3 \theta)^2}{a^2 - r^2cos^2 \theta} + r^2} ~dr ~d \theta
$

Is my solution correct?
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November 7th, 2017, 04:51 PM   #2
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The above integrand is incorrect. I think it should have been the following:

$\displaystyle

\int_{0}^{2 \pi} \int_{0}^{a} \sqrt{ \frac{( r^2 cos^3 \theta - r^2 cos \theta sin^s \theta )^2+(2r^2 cos^2 \theta sin \theta )^2 }{ a^2 - r^2 cos^2 \theta } + r^2 } ~~~dr ~d \theta
$
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