Evaluate the integral $ \ \iint_S F \cdot dS \ $ where $ \ S: \ x^2+y^2+z^2=16 \ $

Question

Evaluate the integral $ \ \iint_S F \cdot dS \ $ where $ \ S: \ x^2+y^2+z^2=16 \ $ and $ \ F(x,y,z)=\left\langle z,y,x \right\rangle \ $

Then verify that,

$$ \iint_S F \cdot dS=\iiint_V div(F) dV \ $$

Answer:

Let $ \ r(\phi,\theta)=\left\langle 4 \sin \phi \cos \theta, 4 \sin \phi \sin \theta, 4 \cos \phi \right\rangle \ $

$ \therefore r_{\phi} \times r_{\theta} =4 \sin \phi \left\langle 4 \sin \phi \cos \theta, 4 \sin \phi \sin \theta, 4 \cos \phi \right\rangle $ Then how to evaluate $ \ \iint_S F \cdot dS \ $ ?

Help me

Answer 1

Easy way:

definitely we have $$\nabla \cdot F=1$$therefore $$\iiint_V\nabla \cdot F dv=\iiint_V dv=\text{the volume of the sphere }V=\dfrac{4}{3}\pi4^3=\dfrac{256\pi}{3}$$Hard way (hint):

consider $ds=R^2\sin\theta d\theta d\phi \hat{a}_r$

Answer 2

It might be easier to start this in Cartesian... you can still convert mid way through

$dS = \frac 14 (x,y,z)\ dx\ dy\\ \iint \frac 14 (z,y,x)\cdot(x,y,z)\ dx\ dy\\ \iint \frac 14 (2xz + y^2)\ dx\ dy$

and then convert to cylindrical

$\iint \frac 14 (2(4\cos\theta\sin\phi)(4\cos\phi) + (4\sin\theta\sin\phi)^2)(16\sin\phi)\ d\phi\ d\theta$

You get to the same place:

$\int_0^{2\pi}\int_0^{\pi} 128\cos\theta\sin^2\phi\cos\phi+ 64\sin^2\theta\sin^3\phi\ d\phi\ d\theta$

Lets flip the order of integration since $\int_0^{2\pi} \cos\theta\ d\theta = 0$ and $\int_0^{2\pi} \sin^2\theta\ d\theta = \pi$

$\int_0^{\pi} 64\pi\sin^3\phi\ d\phi\\ \int_0^{\pi} 64\pi(1-\cos^2\phi)\sin\phi\ d\phi\\ 64\pi(\cos \pi - \frac 13 \cos^3 \phi)|_0^\pi\\ \frac 43(64\pi)$

and by the divergence theorem

$\iiint \nabla F dV = \iiint \nabla dV = \frac 43\pi 4^3$