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$C$ be the curve of intersection of the hemisphere $x^2+y^2+z^2=2ax$ and the cylinder $x^2+y^2=2bx$ , where $0<b<a$ ; how to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$ using Stoke's theorem ? I can't parametrize the curve $C$ nor able to get the surface $S$ . Please help . Thanks in advance

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  • $\begingroup$ Very similar to this one. $\endgroup$
    – Fabian
    Dec 30, 2015 at 12:50
  • $\begingroup$ @Fabian : finding the surface seems more difficult to me :( and even a parametrization $\endgroup$
    – user228169
    Dec 30, 2015 at 12:56
  • $\begingroup$ @Fabian : And could you please add the additional details in the previous answer Please ? $\endgroup$
    – user228169
    Dec 30, 2015 at 12:59

1 Answer 1

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There are at least to ways you can parametrize curve $C$.

Method 1. Since the curve is at the intersection of the sphere and the cylinder, it belongs to the cylinder, which has radius $b$ and is shifted $b$ units along the $x$ axis, so you can write: $$ x=b\cos\theta+b, \quad y =b\sin\theta, \quad 0\le \theta \le 2\pi, $$ but since it also belongs to the sphere, you can write $$ z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(b\sin\theta)^2-(b\cos\theta+b-a)^2}, $$ which gives you a first parametrization.

Method 2. The cylinder has polar equation $r(\theta)=2b\cos\theta$, so $$ x=r(\theta)\cos\theta=2b\cos^2\theta, \quad y =2b\cos\theta\sin\theta, \quad -\pi/2\le \theta \le \pi/2, $$ and again $$ z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(2b\cos\theta\sin\theta)^2-(2b\cos^2\theta-a)^2}. $$ Note the difference of variations of $\theta$ in the two parametrizations.

This being said, Stokes theorem is probably more appropriate for this question, so you are going to need to parametrize the surface. You could do it in spherical coordinates, but I think cartesian coordinates are the way to go here. Take $x$ and $y$ as parameters and you have: $$ x=x, \quad y=y, \quad z=\sqrt{a^2-y^2-(x-a)^2}, $$ where $x$ and $y$ belong to the projection of the cylinder in the $xy$ plane, that is, $$ (x,y)\in D:=\{(r,\theta)\;|\; -\pi/2\le \theta \le \pi/2,0\le r\le 2b\cos\theta \} $$ It follows by the Stokes theorem that your integral equals $$ \iint_D \nabla\times\vec{F}(x,y)\cdot \vec{r_x}\times \vec{r_y}\; dA, $$ where $\nabla\times\vec{F}(x,y)$ is the curl of your field expressed as a field depending only on $x$ and $y$. I am sure you can takes computations from there.

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  • $\begingroup$ I don't get it . Once you're saying Cartesian is the way , then again you are using polar co-ordinates ; what is it ? and how did you get the range for $\theta$ and $r$ ? $\endgroup$
    – user228169
    Jan 4, 2016 at 14:53
  • $\begingroup$ Yes, the parametrization is in cartesian coordinates, as the parameters are $x$ and $y$. But the $D$ is expressed in polar coordinates, so you will have to do the change of variables to finish the integral. The reason why I do this is to simplify $r_x \times r_y$, you can check that it is very easy to compute. You can however parametrize from the start in polar coordinates, it will lead you to the exact same integral. $\endgroup$
    – Kuifje
    Jan 4, 2016 at 15:03
  • $\begingroup$ And how do you get the range for $r$ and $\theta$ ? $\endgroup$
    – user228169
    Jan 4, 2016 at 15:04
  • $\begingroup$ As for the range for $\theta$ and $r$, check Method 2. $D$ is the region inside the curve $C$, so you can use the work done to parametrize $C$. $\endgroup$
    – Kuifje
    Jan 4, 2016 at 15:05

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