# Evaluating $\iint_s \vec F \cdot \hat n ds$ around the curved surface of cylinder cut by plane at $45^\circ$

I need to calculate the surface integral of $F(x,y,z) = \hat i x +\hat j y + \hat k z$ on the curved part of surface $x^2+z^2 = 1, x+y=2,$ and $y$ goes from $1$ to $3$ as shown in following figure. How do I evaluate $\displaystyle \iint_S \vec F .\hat n ds$ this surface?

EDIT::I couldn't do it via parametrization, I got the above figure which is incorrect. Using this formula $\iint_s \vec F \cdot \frac{\nabla \phi }{|\nabla \phi|}\sqrt{1 + (z_x)^2 + (z_y)^2} dx dy$ I got the following. Not sure if it's correct. $$\int_1^3 \;dy \int_{-1}^{2-y} \vec F(x, y , \sqrt{1-x^2})\cdot \frac{x \hat i + \sqrt{1-x^2}\hat k}{\sqrt{1-x^2}} dx \\ + \int_1^3 \;dy \int_{-1}^{2-y} \vec F(x, y , -\sqrt{1-x^2})\cdot \frac{x \hat i - \sqrt{1-x^2}\hat k}{\sqrt{1-x^2}} dx$$

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## 1 Answer

Instead of calculating the integral over this surface, you can calculate it over the two surfaces that would close the shape (the base circle and the slanted circle from the cutting plane) and then negate the result. Over these surfaces, which are flat, recall also that the integral is just the flux of the field passing through the surface.

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that way I can verify my answer ... I need to calculate that surface integral (asked by question) – hasExams May 31 '13 at 12:27
Are you sure? Often the entire point of questions such as these is for you to find an alternative and much simpler way of calculating the answer. Unless the question specifically says by integrating across that surface you should be free to choose any valid method of calculation – john May 31 '13 at 12:29
I could have used divergence theorem (original problem was for closed surface) for the beginning but couldn't as part of exercise ... and while evaluating the surface integral, I got stuck on this part. Any ideas how I can do it other than indirectly evaluating? – hasExams May 31 '13 at 12:36
well your method may work (I would need to look a few things up and can't at the moment) but if you really want to do it literally then I would recommend changing to cylindrical coordinates – john May 31 '13 at 12:43
I tried that, that's where I got that above graph. $(\cos (\theta), \sin (\theta), t)$ but I was having problem with bounds for $\theta$. Check the revision – hasExams May 31 '13 at 12:47