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Evaluate the integral $\iint_S\mathbf{F}\cdot\hat{\mathbf{n}}\;dS$, where $\mathbf{F}=(x,y,1)$ and $S$ is the surface $z=1-x^2-y^2$, for $x^2+y^2\le1$, by two methods.
$\qquad$(a) First, by direct computation of the surface integral.
$\qquad$(b) Second, by using the divergence theorem.

For this question, I can't quite grasp on how to find the unit normal $\hat{\mathbf{n}}$ to do the direct computation. I can see that $G(x,y,z) = 1 - x^2 - y^2 - z$ and thus calculate the $\nabla G$ and then $|\nabla G|$ to find that $\hat{\mathbf{n}} = \nabla G/|\nabla G|$. Is there an easier way to find $\hat{\mathbf{n}}$ without using calculation? Also, what do I integrate with respect to?

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    $\begingroup$ $\mathbf n dS = (-\frac {\partial z}{\partial x}, -\frac {\partial z}{\partial y}, 1) \ dy\ dx$ $\endgroup$
    – Doug M
    Mar 6 '18 at 1:19
  • $\begingroup$ if you look at the surface area integral, you'll see that the division by the norm of the gradient cancels when you are computing flux integrals. $\endgroup$ Mar 6 '18 at 3:02
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I don't know why they don't teach this in school. You have a surface expressed as a vector function of $2$ parameters, here $x$ and $y$. $$\vec r=\langle x,y,z\rangle=\langle x,y,1-x^2-y^2\rangle$$ Then find the differential $$d\vec r=\langle1,0,-2x\rangle dx+\langle0,1,-2y\rangle dy$$ To get the vector element of area you just take the cross of the $2$ vector differentials you got in the last step $$\begin{align}d^2\vec A&=\pm\langle1,0,-2x\rangle dx\times\langle0,1,-2y\rangle dy\\ &=\pm\langle2x,2y,1\rangle dx\,dy\\ &=\langle2x,2y,1\rangle dx\,dy\end{align}$$ Because from the context of the problem we want the upward normal, not the downward one. Then you express the vector field $\vec F$ in terms of the same variables that you used to parameterize the surface, but you already have done so: $$\vec F=\langle x,y,1\rangle$$ Then you get the limits. $x$ can range from $-1$ to $1$ but for a given $x$, $-\sqrt{1-x^2}\le x\le\sqrt{1-x^2}$, so now $$\begin{align}\int\int\vec F\cdot d^2\vec A&=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\langle x,y,1\rangle\cdot\langle2x,2y,1\rangle dy\,dx\\ &=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(2x^2+2y^2+1)dy\,dx\\ &=\int_{-1}^1\frac13(8x^2+10)\sqrt{1-x^2}dx\\ &=\frac13\int_{-\pi/2}^{\pi/2}(8\sin^2\theta+10)\cos^2\theta\,d\theta\\ &=\frac13\left(2\left(\frac{\pi}2\right)+10\left(\frac{\pi}2\right)\right)=2\pi\end{align}$$ On the bottom, $$\begin{align}\vec r&=\langle x,y,0\rangle\\ d\vec r&=\langle1,0,0\rangle dx+\langle0,1,0\rangle dy\\ d^2\vec A&=\pm\langle1,0,0\rangle dx\times\langle0,1,0\rangle dy\\ &=\pm\langle0,0,1\rangle dx\,dy\\ &=-\langle0,0,1\rangle dx\,dy\end{align}$$ Because outward on the bottom means down. Then $$\begin{align}\int\int\vec F\cdot\vec d^2\vec A&=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\langle x,y,1\rangle\cdot\langle0,0,-1\rangle dy\,dx\\ &=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}-1\,dy\,dx\\ &=-\int_{-1}^12\sqrt{1-x^2}dx\\ &=-2\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=-2\left(\frac{\pi}2\right)=-\pi\end{align}$$ Over the whole volume we need $$\vec\nabla\cdot\vec F=1+1+0=2$$ So $$\begin{align}\int\int\int\vec\nabla\cdot\vec Fd^3V&=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{1-x^2-y^2}2\,dz\,dy\,dx\\ &=2\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(1-x^2-y^2)dy\,dx\\ &=2\int_{-1}^1\frac43(1-x^2)^{3/2}dx\\ &=\frac83\int_{\pi/2}^{\pi/2}\cos^4\theta\,d\theta\\ &=\frac83\int_{\pi/2}^{\pi/2}\frac14\left(1+2\cos2\theta+\cos^22\theta\right)d\theta\\ &=\frac23\left(\pi+2(0)+\frac{\pi}2\right)=\pi\end{align}$$ So the statement of the divergence theorem that $$\oint\vec F\cdot d^2\vec A=2\pi-\pi=\int\int\int\vec\nabla\cdot\vec F d^3V=\pi$$ Is confirmed.

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  • $\begingroup$ So i interpreted the surface in the problem as not being closed, it seems you did not. I'm not sure who is right $\endgroup$ Mar 11 '18 at 22:48
  • $\begingroup$ @qbert Since the problem statement required use of the divergence theorem we had to close the surface somehow :) I didn't interpret the curved surface as being closed but we had to compute some closure for the divergence theorem to be applicable. $\endgroup$ Mar 11 '18 at 23:03
  • $\begingroup$ Indeed, I closed the surface as well and then subtracted. Sorry I think I misinterpreted your answer. +1 $\endgroup$ Mar 11 '18 at 23:13
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As the comment mentions, it's easy to find the normal to a surface given by the graph, it's just $$ \nabla g(x,y,z) $$ where $g(x,y,z)=z-f(x,y)$ and $f$ is the graph. Here we have $$ g(x,y,z)=z+x^2+y^2-1\implies \nabla g=(2x,2y,1) $$

With the first method, you are evaluating $$ \int\int_D (f_x,f_y,1)\cdot\vec{F}\mathrm d D $$ with $D$ the unit disc, a picture would be very helpful. We projected down (really they did this for us) to find the $(x,y)$ coordinates over which the relevant portion of the surface lies. This works out to be $$ \int\int_D2x^2+2y^2+1\mathrm dD $$ obviously polar coordinates are the play, where we have $$ \int\int_D2x^2+2y^2+1\mathrm dD=\int_0^{2\pi}\int_0^1 2r^3+r\;\mathrm dr\mathrm d\theta\\ =2\pi $$ Below we verify by the divergence theorem, don't look if you want to tackle this yourself

Now, for the divergence theorem, close up the net by adding in the disc cap (it is easy to see the contribution here is just $-\pi$, since the vector field is constant $1$ in the $z$ coordinate and the surface normal is $(0,0-1)$). Then the divergence theorem gives you the flux through this region as, using polar coordinates for the volume integral and noting that $\nabla\cdot F=2$, $$ \int_{V}\nabla\cdot F\mathrm d V=2\int\int_D 1-x^2-y^2\mathrm d D\\ =2\pi \int_0^1 r(1-r^2)\mathrm dr=\pi $$ subtracting the contribution of the cap, we have our result.

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