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Take the vector field given by: $F= (y^2+yz)i+(\sin(xz)+z^2)j+z^2k$

a) Calculate the divergence, $\operatorname{div}F$.

b) Use the divergence theorem to calculate the flux $$\int_S F\cdot dA $$

through a sphere or radius 2 centered at the origin oriented with an outward pointing unit normal.

For the divergence of $F$, I found it to be $2z$. I'm pretty sure I need to change the integral into spherical coordinates, but I'm not sure if that's right. I'm also not understanding how I would find the limits for the integral as well.

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  • $\begingroup$ The divergence is a scalar. Why does your divergence have a "k" in it? Also, you're missing a nontrivial term from the coefficient of $j$ which has $y$-dependence. That said, the answer is the same with that mistake... $\endgroup$ – James S. Cook Dec 12 '14 at 23:36
  • $\begingroup$ Do you really mean $dA(=dxdy)$? $\endgroup$ – Adam Hughes Dec 13 '14 at 0:02
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A good parametrization for your surface is:

$$\mathbf r(\theta,\phi)=\langle x= 2\sin\phi\cos\theta,y=2\sin\phi\sin\theta,z=2\cos\phi\rangle.$$

Where $\phi$ is the angle between the positive $z$-axis and the vector $\boldsymbol v$ with tail on the origin and tip at a point on the sphere, and $\theta$ is the angle between the positive $x$-axis and the projection of the vector $\boldsymbol v$ onto the $xy$-plane. So your limits of integration would be:

$$0\leq\theta\leq 2\pi,\\0\leq\phi\leq\pi,\\0\leq\rho\leq2.$$

Divergence theorem tells you that:

$$\iint\limits_S \mathbf F \cdot d\mathbf S = \iiint\limits_E \text{div}\mathbf F\,dV.$$

The last triple integral by Fubini is the iterated integral with the bounds I proposed, do change of variables and don't forget the jacobian $\rho^2\sin\phi$.

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    $\begingroup$ note that OP has edited the original problem statement, and $F$ is substantially different now $\endgroup$ – BaronVT Dec 12 '14 at 23:49
  • $\begingroup$ Then would I be multiplying my divF with r(θ,ϕ)=⟨2sinϕcosθ,2sinϕsinθ,2cosϕ) in the integral? $\endgroup$ – lilskatingazn Dec 12 '14 at 23:59
  • $\begingroup$ @lilskatingazn No, your integrand is $2z=2\times2\cos\phi$. When you change variables to spherical coordinates you have to multiply by the jacobian, which will leave you with the integrand: $4\rho^2\cos\phi\sin\phi$. I edited the part of the parametrization to make it clearer for you. $\endgroup$ – Vladimir Vargas Dec 13 '14 at 0:04
  • $\begingroup$ @NotStrang oh that's much clearer. thank you $\endgroup$ – lilskatingazn Dec 13 '14 at 0:10
  • $\begingroup$ @lilskatingazn you're welcome. $\endgroup$ – Vladimir Vargas Dec 13 '14 at 0:10
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The divergence is

$$ \partial_x (y^2 + yz) + \partial_y (\sin(xz) + z^2) + \partial_z (z^2) \\ = 2z.$$

The divergence theorem tells you that the integral of the flux is equal to the integral of the divergence over the contained volume, i.e.

$$ \int_B \nabla \cdot F \,dx\,dy\,dz = \int_B 2z \,dx\,dy\,dz $$

where $B$ is the ball of radius $2$ (i.e. the interior of the sphere in question). If you want to set up limits of integration, use $x^2 + y^2 + z^z = 4$. (and integrating with respect to $z$ first seems fruitful)

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