Question

Suppose f⃗ (x,y,z)=⟨x,y,4z⟩f→(x,y,z)=⟨x,y,4z⟩. let w be the solid bounded by the paraboloid z=x2+y 2 z=x2+y2 and the plane z=9.z=9. let ss be the closed boundary of ww oriented outward. (a) use the divergence theorem to find the flux of f⃗ f→ through s.

Answer 1

$\mathbf F(x,y,z)=\langle x,y,4z\rangle$

By the divergence theorem, the surface integral taken over $S$,

$\displaystyle\iint_S\mathbf F\,\mathrm dS$

is equivalent to the triple integral of the divergence of $\mathbf F$ over $W$, the space bounded $S$,

$\displaystyle\iiint_W(\nabla\cdot\mathbf F)\,\mathrm dV$

We have

$\nabla\cdot\mathbf F=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial(4z)}{\partial z}=1+1+4=6$

The latter integral is then given by

$\displaystyle6\iiint_W\mathrm dV=6\int_{|x|\le3}\int_{|y|\le3}\int_{x^2+y^2\le z\le9}\mathrm dz\,\mathrm dy\,\mathrm dx=648$