How to evaluate surface integral of a cylinder above a square

Question

I'm asked to Evaulate the surface intergral of $$\int_{}^{}\int_{}^{} k \cdot\ \mathrm{d}S\ $$.

where $S$ is part of a cylinder $z = \sqrt {1-x^2}$ that lies above the square with vertices of $(-1,-1), (1,-1), (-1,1), (1,1)$ and has upward orientation.

I'm struggling with the fact that $k$ is alone in the integral. This question is unlike others I've practiced because of the square at the bottom. Should I just assume that it is laying on $z=0$?

How should I handle the square vertices and $k$? I'm struggling on where to even begin this problem. Any help would be appreciated, this is a sample final problem for my calculus class and answer is given, but the work isnot shown.

Answer 1

Consider the half cylinder above $xy$-plane, enclosed by $S$, the bottom square $R$, and two half disks on $y=1$ and $y=-1$. Then try to show that the integral is same as $$ \iint_{R} k\cdot dS $$ by applying the divergence theorem on the half cylinder.

Answer 2

HINT

Let use

• $x=r\cos \theta$
• $z=r\sin \theta$
• $y=y$

and since $r=1$ we have

$$\int_{-1}^{1}dy\int_{0}^{\pi} d\theta$$