MHB Surface Integral of $F$ Over Region V

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The discussion focuses on calculating the surface integral of the vector field F over the region V, defined by a hemisphere and a plane. Two methods are proposed: directly integrating the dot product of F with the outward normal vector n over the surface S, or applying the divergence theorem to simplify the calculation. The divergence of F is calculated as ∇·F = 2z - 2, which can be integrated over the volume of the hemisphere for an easier solution. The participants emphasize the advantages of using the divergence theorem for efficiency. The conversation highlights the mathematical intricacies involved in evaluating the surface integral.
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).
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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