Multivariable Calculus Surface Integral Calculation

Question

I have a surface bounded by $x^2+y^2=1$ and $x^2+y^2=9$ (cylinders) as well as the planes z=0 and z=3.The vector field is $(yx^3,xy^3,x)$. I know this involves the divergence theorem, where I would have to set up find the partial and then the triple integral. I have got the part I need to integrate> I am having a little setting up the bounds for the integral. I know one would def. be from z=0 to z=3, but for the others i am not quite sure. Would I use polar coordinates given the $x^2+y^2$ terms? Any help is appreciated, thank you.

Answer 1

You were probably asked to find $$ \int_{S} \vec{F}\cdot d\vec{S} $$ where $\vec{F}=(yx^{3},xy^{3},x)$, where $S$ is the surface of the solid you described, and where you are using the outward normal on $S$. In that case, the divergence theorem implies $$ \int_{S}\vec{F}\cdot d\vec{S}=\int_{V}\nabla\cdot \vec{F}\,dV=\int_{V}(3yx^{2}+3xy^{2})\,dV, $$ where $V$ is the volume you described. Cylindrical coordinates works great here because the volume you described is $0 \le \theta \le 2\pi$, $1 \le r \le 3$, and $0 \le z \le 3$. Then $x=r\cos\theta$, $y=r\sin\theta$ leads to $$ \int_{S}\vec{F}\cdot d\vec{S}=\int_{0}^{3}\int_{0}^{2\pi}\int_{1}^{3}3r^{3}(\sin\theta\cos^{2}\theta+\cos\theta\sin^{2}\theta)r\,dr\,d\theta\,dz. $$ This integral isn't bad at all because $$ \int \sin\theta\cos^{2}\theta+\cos\theta\sin^{2}\theta\,d\theta=-\frac{1}{3}\cos^{3}\theta+\frac{1}{3}\sin^{3}\theta+C. $$