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Given the surface $S= \left\{(x,,y,z)|z=\sqrt{x^2+y^2},1\leq x^2+y^2\leq2x , 0\leq y \leq x \right\}$ , the surface integral $\iint_S\mathrm z ds\,$ describes the mass when the denisty function is $\rho(x,y,z)=z$ , find the surface mass with parametric form $r(r,\theta)=(x(r,\theta), y (r,\theta) , z(r,\theta))$

My try :

The parametric form would be $r(r,\theta)=<rcos\theta,rsin\theta,r$ and since $ds=||r_r X r_\theta||$ we get $r_r=<cos\theta,sin\theta,1>$ $r_\theta=<-rsin,\theta ,rcos\theta,0>$ and so we get $||r_r X r_\theta||=\sqrt{2}r$ after that I tried to find the projection ( hope the word is right) on x,y axis( but in parametric form) and got $1\leq r\leq 2cos\theta$ and $0 \leq sin\theta \leq cos\theta$ I found $\theta$ bounds and after that I got stuck, I did $sin\theta=0$ and got $0,\pi,2\pi$(because $\theta$ is $0\leq \theta \leq 2\pi $) and for $sin\theta=cos\theta$ I got $\theta= \frac {\pi}{4},\frac{5\pi}{4}$ so my bounds are $0\leq \theta \leq \frac {\pi}{4}$ and $\frac{5\pi}{4} \leq \theta \leq 2\pi$

after that I got stuck , I did not know how to apply all of this on the integral and if what I did is right , appreciate any tips and help!

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1 Answer 1

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Please note that as $x \geq y$ and $y \geq 0, \frac{5\pi}{4} \leq \theta \leq 2\pi$ is not part of the region. Otherwise your working is correct.

Parametrization of the surface is $C(r, \theta) = (r\cos\theta, r\sin\theta, r), 1 \leq r \leq 2 \cos\theta, 0 \leq \theta \leq \frac{\pi}{4}$

$|C_r \times C_{\theta}| = r \sqrt2$

The integral becomes

$\displaystyle \int_0^{\pi/4} \int_1^{2\cos\theta} r \cdot r \sqrt2 \ dr \ d\theta = \frac{1}{36} (80 - 3 \sqrt2 \ \pi)$

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  • $\begingroup$ thank you for the answer , can you please explain why it is not part of the region? I cannot understand it $\endgroup$
    – Adamrk
    May 6, 2021 at 11:47
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    $\begingroup$ Ok if $y \geq 0$, we are in first or second quadrant in 2D so $0 \leq \theta \leq \pi$, correct? $\endgroup$
    – Math Lover
    May 6, 2021 at 11:50
  • $\begingroup$ yes correct , I think it is clear now $\endgroup$
    – Adamrk
    May 6, 2021 at 11:52
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    $\begingroup$ OK. and as $y \leq x$, it cannot be second quadrant (second quadrant has $x$ negative and $y$ positive). $\endgroup$
    – Math Lover
    May 6, 2021 at 11:54

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