MHB Surface Integral of $F$ Over Region V

richatomar
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Let V be the region bounded by the hemisphere z=1-sqrt(1-x^2-y^2) and the plane z=1, and let S be the surface enclosing V. consider the vector field $F= x(z-1)\hat{\imath}+y(z-1)\hat{\jmath}-xy\hat{k}$.
 
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richatomar said:
Let V be the region bounded by the hemisphere z=1-sqrt(1-x^2-y^2) and the plane z=1, and let S be the surface enclosing V. consider the vector field $F= x(z-1)\hat{\imath}+y(z-1)\hat{\jmath}-xy\hat{k}$.

Okay, I'm considering it! What would you like to do with it? Do you want to integrate $\int \vec{F}\cdot \vec{n} dS$? Do you want to integrate directly? Or use the divergence theorem?

Given that $z= 1- \sqrt{1- x^2- y^2}$, $\vec{n}dS= \left(\frac{x}{\sqrt{1-x^2-y^2}}\vec{i}+ \frac{y}{\sqrt{1-x^2-y^2}}\vec{j}+ \vec{k}\right)dxdy$. Also, since $z- 1= \sqrt{1- x^2- y^2}$, $\vec{F}= x(z-1)\vec{I}+ y(z-1)\vec{j}+ xy\vec{k}= x\sqrt{1- x^2- y^2}\vec{i}+ y\sqrt{1- x^2- y^2}\vec{j}+ xy\vec{k}$. Take the dot product of those vectors and integrate over the surface of the hemisphere.

Or, to use the divergence theorem, integrate $\nabla\cdot \vec{F}= 2z- 2$ over the volume of the hemisphere (probably much easier than the previous method).
 
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