# Evaluating a surface integral.

I've been working on this problem for a while now, but the answer I've been getting every time seems to be wrong. At least according to webassign.

Question: Evaluate the surface integral $\iint\limits_s F · dS$ for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. $F(x, y, z) = x^2 \hat i + y^2 \hat j + z^2 \hat k$, S is the boundary of the solid half-cylinder $0 ≤ z ≤ \sqrt{9 − y^2}$, $0 ≤ x ≤ 3$.

I divide it up into 4 parts and get the following:

$S_1:$ $r(x,y)=<x, y,\sqrt{9-y^2}>$ for $0 \le x \le 3, -3 \le y \le 3$.

$\int_0^3 \int_{-3}^{3} \frac{y^3}{\sqrt{9-y^2}} +9-y^2 \,dy\,dx = 108$.

$S_2:$ $x=2$ for $-3 \le y \le 3, 0 \le z \le \sqrt{9-y^2}$.

$\int_{-3}^3 \int_{0}^{\sqrt{9-y^2}} 4 \,dz\,dy = 18\pi$.

For $S_3$ and $S_4$ we get that integrals equal zero. Thus, $\iint\limits_s F · dS = 18\pi+108$.

Is this answer incorrect? Any help is appreciated!

• I think $S_2$ should be at $x=3$, not $x=2$? Have you learnt divergence theorem? Since it is a closed surface, you can actually use divergence theorem to check your answer. – mathshungry Nov 26 '16 at 5:28
• @mathshungry Oh right! I should've noticed, silly me. No, I haven't learned divergence theorem yet. It's next week's lecture, but I'll take a look at it now. Thank you! – Curious Math Student Nov 26 '16 at 21:31