Calculate $\int sze^z dS$ on Unit Sphere

[richatomar]

calculate $\displaystyle \int sze^z dS$
where S is the protion of the unit sphere centered at the origin such that x,y <0, z>0.

 

[HallsofIvy]

calculate $\displaystyle \int sze^z dS$
where S is the protion of the unit sphere centered at the origin such that x,y <0, z>0.

"Portion", not "protion"! I know that "dS" is the surface area integral but what is that small "s"?

If you mean just $\int\int ze^z dS$, then the unit sphere centered at the origin can be written as $x^2+ y^2+ z^2= 1$ or, in parametric equations, $x= cos(\theta)sin(\phi)$, $y= sin(\theta)sin(\phi)$, $z= cos(\phi)$ or, equivalently, as the vector function $(cos(\theta)sin(\phi), sin(\theta)sin(\phi), cos(\phi))$. The "differential of surface area" of the unit sphere is given by $sin(\phi)d\theta d\phi$. The region such that x and y are both negative and z is positive is that where $\theta$ is from $\pi/2$ and $3\pi/2$ and z is from 0 to $\pi/2$. $ze^z$, with $z= cos(\phi)$ is $cos(\phi)e^{cos(\phi)}$ so the integral is $\int_0^{\pi/2}\int_{\pi/2}^{3\pi/2} sin(\phi)cos(\phi)e^{cos(\phi)} d\theta d\phi$.