# Finding surface mass with integrals

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 $$r(r,\theta)= and since $$ds=||r_r X r_\theta||$$ we get $$r_r=$$ $$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!

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

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

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