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I try to calculate this surface integral, however my method does not work. Problem

Attempted sol: I tried to project the surface at the $xy$ plane using this formula and the relationship $z = \sqrt{x^2-y^2}$. Projection of a surface However, then I got some complicated expression that I do not now what to do with (and how to set up the limits).

cmk
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Curtis
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  • Your projection is not good, because it is not bijective (for example, the points (2,0,2) and (2,0,-2) are at the cone and have the same projection on the xy-plane). The axis of the cone is the $x$-axis, so you need to do the projection at the yz-plane. – André Porto May 27 '19 at 23:47

1 Answers1

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Using a cylindrical substitution,

$$x=\rho\\y=\rho\cos\theta\\z=\rho\sin\theta$$

with bounds $2\le\rho\le3$, and $0\le\theta\le2\pi$. Letting this $(x,y,z)$ be the value of $f(\rho,\theta)$, the integral becomes $$\int_0^{2\pi}\int_2^3\rho^4\cos^2\theta\left|\left|\frac{\mathrm df}{\mathrm d\rho}\times\frac{\mathrm df}{\mathrm d\theta}\right|\right| \,\mathrm d\rho\mathrm\, d\theta$$ which is simply $$\int_0^{2\pi}\int_2^3\rho^4\cos^2\theta\sqrt{2}\rho \,\mathrm d\rho\mathrm\, d\theta$$ or $$\frac{665}{6}\sqrt{2}\int_0^{2\pi}\cos^2\theta \,\mathrm d\theta=\frac{665}{6}\sqrt{2}\pi$$

auscrypt
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