4
$\begingroup$

I have 2 questions on stokes and divergence theorem each. I think I have done both and I just want to make sure that I did them correctly.

Question 1
Let $C$ be the boundary of the surface $S={(x,y,z):z=4-x^2-y^2,x^2+y^2\le4, x,y\ge0}$ with orientation related by the right-hand rule to the upward orientation of $S$. For $E(x,y,z)=[3yz,zx,2xy]^T$, apply Stoke's Theorem to calculate the closed line integral $\int E\cdot Tds$

My Work
So by Stokes' theorem I know $$\int E\cdot Tds = \iint(\nabla \times E)\cdot dS$$ where $dS=[-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1]$. So $\nabla \times E=[2x,0,-2x]$ and $dS=[-2x,-2y,1]$ which leads to the double integral $$\iint -4x^2-2x \mathrm{d}x\mathrm{d}y$$ but we can change this into polar coordinates. So we have $$\int_0^{2\pi}\int_0^2 (-4(r\cos\theta)^2-2(r\cos\theta))rdrd\theta$$ which equals $-\frac{32\pi}{3}$.

Question 2
Apply the Divergence Theorem to calculate the normal surface integral of $E$ over $S$, where $S$ is the sphere of radius 3 centered at $(0,0,0)$ oriented outward and $E(x,y,z)=[x^3+yz,y^3+zx,z^3+xy]^T$

My Work
Divergence theorem says $$\iint E\cdot \vec ndr=\iiint(\nabla \cdot E)dxdydz.$$ So the $\nabla \cdot E = 3(x^2+y^2+z^2)$ which screams spherical coordinates. So my triple integral is now $$\int_0^3\int_0^{2\pi}\int_0^\pi (3p^2)p^2\sin(\phi)d\phi d\theta dp= \frac{2916\pi}{5}$$

Are both of those correct?

$\endgroup$
3
  • $\begingroup$ For your second problem, if you're integrating $\phi$ from $0 \rightarrow \pi$ how do you still have a $\phi$ in the final answer? $\endgroup$ Dec 11, 2012 at 1:53
  • $\begingroup$ @MSEoris sorry i meant to type \pi but typed \phi. Typo, fixed now. $\endgroup$
    – Richard
    Dec 11, 2012 at 1:54
  • $\begingroup$ Changed spelling Stokes's $\endgroup$
    – GEdgar
    Dec 11, 2012 at 1:58

1 Answer 1

3
$\begingroup$

This is a community-wiki answer trying to remove this question from the unanswered queue.


Question 1:

$curl(E)=[2x,0,−2x].$

This is not correct. If $\mathbf{E} = (3yz,zx,2xy)$, then $$ \nabla \times \mathbf{E} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x &\partial_y & \partial_z \\ 3yz & zx & 2xy \end{vmatrix} = (2x-x, 3y-2y, z-3z) = (x,y,-2z). $$ Hence restricted on this surface $z=4-x^2-y^2$ we have: $$ \nabla \times \mathbf{E}\big\vert_{S} = (x,y, 2x^2+2y^2-8). $$

$ds=[-2x,-2y,1]$.

There is a minor sign error in this, notice $$ d\mathbf{S} = \left(-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1\right) = (2x,2y,1). $$

Double integral $\iint(-4x^2-2x ) \,dxdy$ into polar coordinates $\int_0^{2\pi}\int_0^2 (-4(r\cos\theta)^2-2(r\cos\theta))rdrd\theta$.

$x,y\geq 0$, and $x^2+y^2\leq 4$, hence the surface is defined in the first octant , and the range for $x,y$ should be the first quadrant in the $xy$-plane. The range of the polar angle should be from $0$ to only $\pi/2$. To combine all the results above: $$ \oint_C \mathbf{E}\cdot \mathbf{t} \,ds = \iint_S\nabla \times \mathbf{E}\cdot d\mathbf{S} \\ = \iint_{\{x,y\geq 0 : \; x^2+y^2\leq 4\}} (2x^2+2y^2 + 2x^2+2y^2-8 ) \,dxdy \\ =\int^{\pi/2}_0 \int^2_0 (4r^2-8)r\,drd\theta = 0.$$


Question 2:

So my triple integral is now $\int_0^3\int_0^{2\pi}\int_0^\pi (3p^2)p^2sin(\phi)d\phi d\theta dp$ = $\frac{2916\pi}{5}$.

The polar integral set up is perfect, just I suggest use $r$ as the radial variable: $$ \int_0^3\int_0^{2\pi}\int_0^\pi (3r^2)r^2\sin \phi \,d\phi d\theta dr. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .