November 4th, 2017, 05:22 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 179 Thanks: 2  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? 
November 7th, 2017, 03:51 PM  #2 
Senior Member Joined: Jan 2017 From: Toronto Posts: 179 Thanks: 2 
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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integral, problem, surface 
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