May 31st, 2017, 05:08 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Line Integral
Let $V$ be the region which is common to the solid sphere $\displaystyle x^2+y^2+z^2 \leq 1 $ and the solid cylinder $\displaystyle x^2+y^2 \leq 0.5. $ Let $\displaystyle dv $ be the boundary of $V$ and $n$ be the unit outward normal drawn at the boundary. Let $\displaystyle F=(y^2+z^2)i+(z^2+x^2)j+(x^2+y^2)k$. Then the value of $\iint_{dv} F.nds $ is equal to A) 0 B) 1 C)1 D) $\displaystyle \pi $ I see that $div F$ is $0$ using Gaussian Divergence method. So how can I proceed further ? Is the answer is option A ? Please help! Thank you 
May 31st, 2017, 06:57 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,653 Thanks: 681 
Since V is three dimensional, its 'boundary' is the surface consisting of cylindrical surface from to and the surface of the sphere above and below that. Do you mean that surface integral, not a line integral? If so, on the cylindrical part of the surface, use parameters and z so that , with going from 0 to and z from to . On the spherical "caps" use spherical coordinates, with , as parameters. , , with going from 0 to and going from 0 to . 

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