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I need to calculate the following surface integral:$\iint x^2dydz+y^2dxdz+z^2dxdy$ on the region $S:(x-a)^2+(y-b)^2+(z-c)^2=R^2$ in the direction of outside normal. So, first i found that normal has coordinates $(\frac{x-a}{R},\frac{y-b}{R},\frac{z-c}{R})$.(I am not sure should I use $+$ or $-$ outside the bracket.) Then, I am calculating $<(x^2,y^2,z^2),(\frac{x-a}{R},\frac{y-b}{R},\frac{z-c}{R})>$ and get $\frac{x^2(x-a)+y^2(y-b)+z^2(z-c)}{R}$. Using spherical coordinates $x=a+R\cos\phi\sin\theta, y=b+R\sin\phi\sin\theta, z=c+R\cos\theta$ I am proceeding with calculating this integral with bounds $0\le\phi\le2\pi, 0\le\theta\le\pi$. Is this correct? So, my only doubts is should I use + or -, and is the following at attempt correct?

Trevor
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  • Make a draw, take a point on the sphere (for example $(x,y,z)=(a,b,R+c)$). Then see if your normal vector at this point is inside or outside the sphere. – Surb Jan 29 '21 at 09:53
  • Outside right, so should it be -? – Trevor Jan 29 '21 at 11:24
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    What you are doing is correct. There is nothing wrong with the sign. You can alternatively apply divergence theorem. – Math Lover Jan 29 '21 at 11:39

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Alternatively apply divergence theorem.

$\displaystyle \iint_S x^2dydz+y^2dxdz+z^2dxdy = \iiint_V (div \vec{F}) \ dV$

$\vec{F} = (x^2,y^2,z^2)$

$div(\vec{F}) = \nabla \cdot \vec{F} = 2(x+y+z)$

Please use $x = a + \rho \cos \theta \sin\phi, y = b + \rho \sin\theta \sin\phi, z = c + \rho \cos\phi$ for the integral.

Math Lover
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  • OK, and is my solution correct if I would go without divergence theorem? And how did you got $div(F)$, shouldn't gradient be $(2(x-a),2(y-b),2(z-c))$? – Trevor Jan 29 '21 at 11:39
  • Just saw you answered my first question... – Trevor Jan 29 '21 at 11:40
  • Divergence of the vector field is $\frac{\partial (x^2)}{\partial x} + \frac{\partial (y^2)}{\partial y} + \frac{\partial (z^2)}{\partial z}$ as the vector field is $(x^2, y^2, z^2)$. – Math Lover Jan 29 '21 at 11:43
  • Right. And one more question:Is this problem equivalent with find flux of vector field $(x^2,y^2,z^2)$ through this sphere? – Trevor Jan 29 '21 at 11:49
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    Yes that is correct. We are finding flux through the given sphere of the vector field $(x^2, y^2, z^2)$. – Math Lover Jan 29 '21 at 11:50
  • Yes, I also think this example is easier using divergende theorem, when it comes to solve the integral etc. – Trevor Jan 29 '21 at 11:53
  • Bounds for integral are $0\le \phi\le 2\pi, 0\le \theta\le \pi, 0\le r\le R$? – Trevor Jan 29 '21 at 11:56
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    Btw I notice that part of the integral will be zero so it should simplify. – Math Lover Jan 29 '21 at 11:57
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    Yes your notation for $\theta$ and $\phi$ is opposite of what I have used in my solution so yes those are correct bounds. – Math Lover Jan 29 '21 at 11:58
  • Yes, yes just saw that. Thanks a lot! – Trevor Jan 29 '21 at 11:59
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    Last bit, given the symmetry of $(x-a), (y-b), (z-c)$ wrt the center of the sphere $(a, b, c)$, that part of the integral would be zero. So the answer should come to $2 (a+b+c) V$. You can do the integral or you know the volume of the sphere is $\frac{4}{3} \pi R^3$. So just use that. – Math Lover Jan 29 '21 at 12:13