Want hint to find surface integral of hemisphere **Let S denote the hemisphere $x^2+y^2+z^2=1$ , $z\ge0$, and let F(x,y,z)=xi+yj. Let n be the unit outward normal of S. Compute the value of the surface integral $\iint_S{F.n} dS$ , using 
a) the vector representation r(u,v)= sinu cosv i+ sinu sinv j + cosu k
b) the explicit representation z=$\sqrt{1-x^2-y^2}$**
Now I have mentioned the formula for surface integral in the picture but have no idea to proceed. How shall I find P[r(u,v)] ,etc? Or  am I using wrong formula? 
I am using textbook T.M Apostol: Calculus, Vol II (2nd edition) 
 A: You are using the correct book formula, but you now have to let correspond the variables in that formula to your situation. I'd write the formula in this way:
$$\int_S{\bf F}\cdot{\bf n}\>{\rm d}\omega=\int_T {\bf F}\bigl({\bf r}(u,v)\bigr)\cdot\bigl({\bf r}_u(u,v)\times{\bf r}_v(u,v)\bigr)\>{\rm d}(u,v)\ ,$$
up to sign. This means that you have to plug in the parametric representation ${\bf r}(u,v)$ of $S$ into the definition of the vector field ${\bf F}$.
The example at hand is rotationally symmetric with respect to the $z$-axis. This fact permits  a considerable simplification of the computation. Note that at each point ${\bf r}\in S$ the field vector ${\bf F}({\bf r})$ is the projection ${\bf r}'$ of ${\bf r}$ to the $(x,y)$-plane. When the geographical latitude of ${\bf r}$ is $\theta$ then $${\bf F}({\bf r})\cdot{\bf n}={\bf r}'\cdot{\bf r}=\cos^2\theta\qquad\left(0\leq\theta\leq{\pi\over2}\right)\ .$$
The area ${\rm d}\omega$ of the infinitesimal latitude annulus defined by $[\theta,\theta+d\theta]$ is given by $2\pi\cos\theta \>d\theta$. We therefore obtain
$$\int_S{\bf F}\cdot{\bf n}\>{\rm d}\omega=2\pi\int_0^{\pi/2}\cos^3\theta\>d\theta={4\pi\over3}\ .$$
