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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?

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  • $\begingroup$ is there a typo here? $x+y=2$ is a superfluous bound to the one given for the cylinder and $z$ is unconstrained. $\endgroup$ – qbert Nov 14 '17 at 1:18
  • $\begingroup$ @qbert yeah sorry. The cylinder was supposed to be $x^2 + z^2$ = 1. I've fixed it $\endgroup$ – Vinny Chase Nov 14 '17 at 1:22
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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 $$ 2\int_0^{2\pi}\int_0^1\int_0^{2-2r\cos \theta}rdydrd\theta\\ =2\int_0^{2\pi}\int_0^1(2r-r^2\cos \theta)dydrd\theta\\ =2\int_0^{2\pi}(1-1/3\cos \theta)d\theta\\ =4\pi $$

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  • $\begingroup$ We haven't done divergence theorem yet. I think thats in two lectures $\endgroup$ – Vinny Chase Nov 14 '17 at 1:45

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