2

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?

enter image description here

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. $$\int_1^3 \;dy \int_{-1}^{2-y} \vec F(x, y , \sqrt{1-x^2})\cdot \frac{x \hat i + \sqrt{1-x^2}\hat k}{\sqrt{1-x^2}} dx \\ + \int_1^3 \;dy \int_{-1}^{2-y} \vec F(x, y , -\sqrt{1-x^2})\cdot \frac{x \hat i - \sqrt{1-x^2}\hat k}{\sqrt{1-x^2}} dx$$

hasExams
  • 2,285

1 Answers1

1

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.

john
  • 5,633
  • 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
  • they are a little tricky. They're based off where the cylinder intersects the plane for each value of y. Given this value, you can determine x and hence the two values of z for these intersection points and from the values of x and z you know the bounds for $\theta$ – john May 31 '13 at 12:57
  • could you give me exact answer for the bounds of $\theta$, if I could fix my above graph, I'll give you (+1) and accept the answer. – hasExams May 31 '13 at 12:59
  • $\theta$ should range from $\theta_0 = \tan^{-1}( \frac{\sqrt{(3-y)(y-1)}}{2-y})$ to $2 \pi - \theta_0$ – john May 31 '13 at 13:11
  • Did I do anything wrong? i used to code ParametricPlot3D[{Cos[x] , t, Sin[x]}, {t, 1, 3}, {x, ArcTan[(3 - t) (t - 1)/(2 - t)], 2 \[Pi] - ArcTan[(3 - t) (t - 1)/(2 - t)]}] and the picture like this pops up. Anyway thank you for your time and effort. this is something that is still amiss. – hasExams May 31 '13 at 13:19
  • you forgot the square root – john May 31 '13 at 13:27
  • Still not getting ParametricPlot3D[{Cos[\[Theta]] , Sin[\[Theta]], t}, {t, 1, 3}, {\[Theta], ArcTan[Sqrt[(3 - t) (t - 1)/(2 - t)]], 2 \[Pi] - ArcTan[Sqrt[(3 - t) (t - 1)/(2 - t)]]}] I am getting this – hasExams May 31 '13 at 13:34
  • the square root is for the numerator only – john May 31 '13 at 13:36
  • looks like this is linear, but this does not go beyond $\pi/2$ – hasExams May 31 '13 at 14:11
  • it's not linear and $theta_0$ has range from 0 to $\pi$ if you choose the appropriate quadrant for arctan – john May 31 '13 at 14:22
  • I got this by splitting it like this $t = 1\to 2$, $\theta = \theta_0 \to 2\pi - \theta_0 $, and $t = 2\to 3$, $\theta = -\theta_0 \to \theta_0 $ gives the upper half. Could you update your answer about how you got that? – hasExams May 31 '13 at 14:36
  • intersection points are where $x=2-y$ hence $(2-y)^2 + z^2 =1$ hence $z=\pm \sqrt{1-(2-y)^2} = \pm \sqrt{(3-y)(y-1)}$ then $\theta = \tan^{-1} (\frac{z}{x})$ – john May 31 '13 at 14:41