Calculate the surface integral of $A:=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2+z^2\leq 4, x\leq0,z\leq0\}$ using a suitable parametrization and the corresponding surface element.

I think this set is a quartersphere, so parametrization is

$\vec f_r:[0,\frac{\pi}{2}]\times[0,\pi] \rightarrow \mathbb{R}^3, (\sigma,\varphi)\mapsto \begin{pmatrix}r \cos(\varphi)\sin(\sigma)\\r \sin(\varphi)\sin(\sigma)\\r \cos(\sigma)\end{pmatrix}, 0\leq r\leq2$

So surfacce integral is

$\int_{\sigma=0}^{\frac{\pi}{2}} \int_{\varphi=0}^\pi 1 \sin(\sigma)d\varphi d\sigma=...=\pi r^2$

Since we're calculating the surface, $r=2$, so the value of this surface integral is $4 \pi$.

Could somebody please tell me, if this is correct?

  • $\begingroup$ So the correct parametrization would be $\vec f_r:[0,\frac{\pi}{2}]\times[0,\frac{\pi}{2}] \rightarrow \mathbb{R}^3 $? Then the value of this integral would be $2 \pi$ i think. $\endgroup$ – fear.xD Jan 26 '15 at 19:26
  • $\begingroup$ I had said something wrong in a previous comment which I deleted. I have now answered the question with the relevant observations and an alternative solution. $\endgroup$ – Git Gud Jan 27 '15 at 0:13

Your parametrization isn't correct, for instance $\vec f_2\left(\frac{\pi}{4}, \frac{\pi}{4}\right)=(1,1,\sqrt{2})\not \in A$. However, due to the obvious symmetry of the solid, your result will still be correct, if you didn't make any mistakes.

To actual parametrize the boundary of $A_r$, where $r>0$ and $$A_r=\{(x,y,z)\in \mathbb{R}^3\colon x^2+y^2+z^2\leq 4 \land x\leq 0 \land z\leq 0\},$$ take $\vec f_r\colon \left[-\frac \pi 2, 0\right]\times \left[\frac \pi 2, \frac{3\pi}2\right]\to \mathbb R^3, (\theta,\varphi)\mapsto r(\cos(\varphi)\sin(\theta), \sin(\varphi)\sin(\theta), \cos(\theta))$.

I'll be confirming your result using a different parametrization.

It's easy to see that $$\partial A_r=\left\{(x,y,z)\in \mathbb R^3\colon -r\leq x\leq 0\land -r\leq y\leq r\land z=-\sqrt{r^2-x^2-y^2}\right\}.$$

Following the notation of this answer, one sets $$\vec r\colon ]-r,0[\times ]-r,r[\to \mathbb R^3, (x,y)\mapsto \left(x,y,f(x,y)\right),$$ where $$f\colon ]-r,0[\times ]-r,r[\to \mathbb R^3, (x,y)\mapsto\left(x,y,-\sqrt{r^2-x^2-y^2}\right).$$

The desired area then comes $$\iint \limits_{]-r,0[\times ]-r,r[}\sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial y}(x,y)\right)^2}\mathrm dx\mathrm dy.$$

In fact $$ \begin{align} \iint \limits_{]-r,0[\times ]-r,r[}\sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial y}(x,y)\right)^2}\mathrm dx\, \mathrm dy&=\iint \limits_{]-r,0[\times ]-r,r[}\sqrt{1+\dfrac{x^2+y^2}{r^2-x^2-y^2}}\mathrm dy\, \mathrm dx\\ &=\iint \limits_{]-r,0[\times ]-r,r[}\sqrt{\dfrac{r^2}{r^2-x^2-y^2}}\mathrm dy\, \mathrm dx\\ &=r\int \limits_{\pi/2 }^{3\pi /2} \int \limits_{0}^r\dfrac{\rho}{\sqrt{r^2-\rho^2}}\mathrm d\rho\,\mathrm d\theta\\ &=-\dfrac{r\pi}{2} \int \limits_{0}^r\dfrac{-2\rho}{\sqrt{r^2-\rho^2}}\,\mathrm d\rho\\ &=-r\pi \Biggl.\sqrt{r^2-\rho^2}\Biggr|_0^r\\ &=\pi r^2. \end{align} $$


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