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Here's the problem: Let $G(x,y,z)= \langle x+\cos z,y+y\sin x,z+\cos y-z\sin x\rangle$. Compute the surface integral $\int\int_SG\cdot ds$ where $S$ is the boundary of the solid which is inside the cylinder $x^2+y^2=4$ bounded above by the plane $z=x$ and below by the $xy$-plane.

First, I got $\mathrm{div} G=3$ Then my integral was structured as so: \begin{eqnarray*} \int_{-\pi/2}^{\pi/2}\int_0^2\int_{-r\cos\theta}^0 3r~\mathrm dz~\mathrm dr~\mathrm d\theta &=& \int_{-\pi/2}^{\pi/2}\int_0^2 3r^2\cos\theta~\mathrm dr~\mathrm d\theta \\ \\ &=& [2^3][\sin(\pi/2)-(-\sin(\pi/2))] \\ \\ &=&2^3\cdot2=16 \end{eqnarray*}

Is this correct, or am I missing something?

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  • $\begingroup$ Well, I believe the divergence theorem states $\int_V \partial_i Q_i dV = \iint Q_i n_i dA$ where $n$ is the outward normal. I think this a vector equation, but may still be applied to your problem, I would recommend putting it in the proper form first. Additionally, your limits on $\theta$ might need to go from $-\pi$ to $\pi$ instead of $-\pi/2$ to $\pi/2$. $\endgroup$ – Charles Apr 23 '16 at 1:40
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Yep, you're answer is completely right. Usually however, the divergence theorem is used when you don't want to explicitly solve the integral $\iint_S G \cdot ds$. In this case, you would say $$\iint_S G \cdot ds = \iiint_V \mbox{div}(G) \, dV $$ You were correct in finding $$\mbox{div}(G)=\bigg(\frac{\partial}{\partial x}x+\cos(z)\bigg)+\bigg(\frac{\partial}{\partial y}y+y\sin(x)\bigg)+\bigg(\frac{\partial}{\partial z}z+\cos(y)-z\sin(x)\bigg)=3$$ So our equation now becomes $$\iint_S G \cdot ds = \iiint_V 3 \, dV$$ It's easier to write the bounds in cylindrical coordinates, so doing that (and remembering to include the extra $r$ term) yeilds \begin{aligned} \iint_S G \cdot ds &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} \int_{0}^{r\cos(\theta)} 3r \, dz \,dr \, d\theta \\ &=16 \end{aligned} Which is what you concluded from your work.

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