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KIM [24]
3 years ago
7

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

the sphere x2 + y2 + z2 = 9. (Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x2 + y2 ≤ 9, oriented downward, and S2 = S1 ∪ S.)
Mathematics
1 answer:
GenaCL600 [577]3 years ago
5 0

Close off the hemisphere S by attaching to it the disk D of radius 3 centered at the origin in the plane z=0. By the divergence theorem, we have

\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV

where R is the interior of the joined surfaces S\cup D.

Compute the divergence of \vec F:

\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2

Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}

So the volume integral is

\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5

From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

Parameterize D by

\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

where 0\le u\le 3 and 0\le v\le2\pi. Take the normal vector to be

\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k

Then taking the dot product of \vec F with the normal vector gives

\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v

So the contribution of integrating \vec F over D is

\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==>  486π/5 - (-81π/4) = 2349π/20

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