# Parameterization for this surface integral

The problem is:

Evaluate the $\int \int_SF* dS$ for the given vector field F and the oriented surface S. for closed surfaces, use the positive (outward) orientation. F(x,y,z) = xi +yj +5k. S is the boundary of the region enclosed by the cylinder $x^2 + z^2$ = 1 and the planes y = 0 and x + y = 2.

I get that these are three separate surfaces. The only one I'm having trouble with is the lateral portion of the cylinder. In my book, they parametrize it as x = sin(t), y=y, z=cos(t). Why do they have x as sin and z cos? Is there a way to have x as cos and z as sin and just change the bounds?

• is there a typo here? $x+y=2$ is a superfluous bound to the one given for the cylinder and $z$ is unconstrained. – qbert Nov 14 '17 at 1:18
• @qbert yeah sorry. The cylinder was supposed to be $x^2 + z^2$ = 1. I've fixed it – Vinny Chase Nov 14 '17 at 1:22

Why not use the divergence theorem? It is particularly easy in this case since $$\nabla\cdot F=2$$ Reducing your surface integral to $$2\int\int\int_V dzdxdy=2\int_0^1\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_0^{2-x}dydxdz$$ where $V$ is the volume described. Note that this is easier in cylindrical coordinates (how the solution is proceeding), where we allow $\theta$ to sweep out the full cylinder, the radius is from $0$ to $1$, and your integral is $$2\int_0^{2\pi}\int_0^1\int_0^{2-2r\cos \theta}rdydrd\theta\\ =2\int_0^{2\pi}\int_0^1(2r-r^2\cos \theta)dydrd\theta\\ =2\int_0^{2\pi}(1-1/3\cos \theta)d\theta\\ =4\pi$$