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 May 31st, 2017, 05:08 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 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: 3,165 Thanks: 867 Since V is three dimensional, its 'boundary' is the surface consisting of cylindrical surface $x^2+ y^2= 0.5$ from $z= -\sqrt{0.5}$ to $z= \sqrt{0.5}$ 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 $\theta$ and z so that $x= \sqrt{0.5}cos(\theta)$, $y= \sqrt{0.5}sin(\theta)$ with $\theta$ going from 0 to $2\pi$ and z from $-\sqrt{0.5}$ to $\sqrt{0.5}$. On the spherical "caps" use spherical coordinates, with $\rho= 1$, as parameters. $x= cos(\theta)sin(\phi)$, $y= sin(\theta)sin(\phi)$, $z= cos(\phi)$ with $\theta$ going from 0 to $2\pi$ and $\phi$ going from 0 to $cos^{-1}(\sqrt{0.5})$.

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