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Studying for the GRE, haven't seen a problem like this for a while:

What is the value of the flux of the vector field $\textbf{F}$, defined on $\mathbb{R}^3$ by $\textbf{F} (x, y, z) = x\textbf{i} + y\textbf{j} + z\textbf{k}$, through the surface $z=\sqrt{ 1 + x^2 + y^2}$ oriented with upward-pointing normal vector field? (Note: $\textbf{i}$, $\textbf{j}$, and $\textbf{k}$ are the standard basis vectors in $\mathbb{R}^3$. )

We've tried using $\int\mathbf{F} \cdot d\mathbf{A}$ and polar coordinates, I'm not sure where we went wrong.

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  • $\begingroup$ Why do you think it went wrong? Do you have the answer? You got something different? $\endgroup$ – DonAntonio Mar 27 '17 at 20:01
  • $\begingroup$ The answer is $2\pi$, but I can't figure out how to get there. $\endgroup$ – setholopolus Mar 27 '17 at 20:06
  • $\begingroup$ @DonAntonio I get a hyperboloid $\endgroup$ – Rafa Budría Mar 27 '17 at 20:43
  • $\begingroup$ @Rafa Of course, you're right! I got the signs messed up...hehe. Thanks! $\endgroup$ – DonAntonio Mar 27 '17 at 20:45
  • $\begingroup$ @DonAntonio, the integral seems to diverge. Maybe a limiting range is needed... $\endgroup$ – Rafa Budría Mar 27 '17 at 20:46
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Parametrize the cone with \begin{cases} x=x\\ y=y \\ z=\sqrt{1+x^2+y^2} \end{cases} with $(x,y) \in \mathbb{R}^2$. The flux equals \begin{align} &\iint_{\mathbb{R}^2} \pmatrix{x\\y\\\sqrt{1+x^2+y^2}}\cdot \pmatrix{1\\0\\\frac{x}{\sqrt{1+x^2+y^2}}} \times \pmatrix{0\\1\\\frac{y}{\sqrt{1+x^2+y^2}}}\; dxdy \\&= \iint_{\mathbb{R}^2} \pmatrix{x\\y\\\sqrt{1+x^2+y^2}}\cdot \pmatrix{\frac{-x}{\sqrt{1+x^2+y^2}}\\\frac{-y}{\sqrt{1+x^2+y^2}}\\1}\; dxdy \\& =\iint_{\mathbb{R}^2} \frac{-x^2}{\sqrt{1+x^2+y^2}}-\frac{y^2}{\sqrt{1+x^2+y^2}}+\sqrt{1+x^2+y^2}\; dxdy \\&= \iint_{\mathbb{R}^2} \frac{-x^2-y^2+(1+x^2+y^2)}{\sqrt{1+x^2+y^2}}\; dxdy \\ & = \int_0^{2\pi}\int_0^{\infty} \frac{r}{\sqrt{1+r^2}}\; dr d\theta \\ & = \lim_{R\rightarrow \infty} 2\pi (\sqrt{R^2+1}-1) = \infty \end{align}

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  • $\begingroup$ Maybe is a good problem try to find some simple contour to get $2\pi$, so is $R=\sqrt{3}$ $\endgroup$ – Rafa Budría Mar 28 '17 at 13:40
  • $\begingroup$ Is it possible that OP forgot to mention something in the question, e.g., "find the flux through the surface bounded by the cone and the plane $z=2$" ? $\endgroup$ – Kuifje Mar 28 '17 at 13:58
  • $\begingroup$ Yes. In that case, maybe the flux through the plane must be included. Who knows, so many OP's forgot the question once the homework expired... $\endgroup$ – Rafa Budría Mar 28 '17 at 14:02

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