# Surface integral of a pyramid in a sphere

Calculate surface integral: $$I = \iint \limits_{S} \frac{xy}{x^2 + y^2} d{S}$$ where S is surface determined by sides of pyramid $x + y + z = 1 \phantom\ (x, y, z \ge 0 ),$ inside sphere $(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 + z^2 \le \frac{1}{4}.$

How should I parameterize the surface $S$? I am also having a question regarding the case of the flux integral: How should I find the unit normal vector of the surface $S$?

The parts of the surface that give a non-zero contribution to the integral are

1) the one over the face $x + y + z = 1$

2) the one over the face $z=0$.

(on the other two faces $x\cdot y=0$ and therefore $f(x,y)=0$).

As regards $I_1$, we have that $dS=\sqrt{3}dxdy$ and $$I_1=\sqrt{3}\iint_{D_1}\frac{xy}{x^2 + y^2}\,dxdy$$ where $D_1=\{(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 +(1-x-y)^2\le \frac{1}{4},y\geq 0,x+y\leq 1\}$.

For $I_2$, $dS=dxdy$ and $$I_2=\iint_{D_2}\frac{xy}{x^2 + y^2}\,dxdy.$$ where $D_2=\{(x - \frac{1}{2})^2 + (y - \frac{1}{3})^2 \le \frac{1}{4},y\geq 0,x+y\leq 1\}$.

• In a case the function was $f(x, y, z) = \frac{x}{x^2 + y^2}$, how would I then calculate the integral on coordinate planes? Also, could you further explain why the integral has value 0 on $z = 0$ plane? Thank you May 23, 2017 at 17:35
• You are right. There is another contribution over $z=0$. This integral is quite messy! May 23, 2017 at 17:37
• @Nemanja Beric I am sorry but I can't find an easier way to do it. Are you supposed to find explicitly the result? May 23, 2017 at 17:49
• Yes we are. It was a problem on an exam. You have already helped me a lot. No, no more questions. Thank you. I thought the $D_1$ should be projected on $z = 0$ plane... May 23, 2017 at 17:56
• I see. $I_1$ should be ok now. For $I_1$ one can try to write the integral with respect to $dydz$ or $dxdz$. Maybe it is easier. May 23, 2017 at 18:16