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After several years I suddenly need to brush up on surface integrals. Looking through my old Calculus book I have been attempting to solve some problems, but the following problem has really made me hit a wall, even though it probably is quite easy to solve:

Find $\int \int_S y dS$, where $S$ is part of the cone $z=\sqrt{2(x^2 + y^2)}$ that lies below the plane $z=1+y$.

So far I have found that $dS = \sqrt{3}$, which then means I have to solve the integral:

$$\sqrt{3}\int \int_S y dx dy$$.

However, I am really stuck on how to proceed from here. I have tried looking at the intersection between the cone and the plane, and transforming the integral to polar coordinates, but can't seem to get anywhere. If someone can help me out a bit here, then I would greatly appreciate it!

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  • $\begingroup$ Can you show your work on polar coordinates? As far as I remember, it is a correct way to go. $\endgroup$ – Niki Di Giano Aug 3 '18 at 19:06
  • $\begingroup$ So, in polar coordinates I use that $y=r \sin(\theta)$. I am, however, unable to see what the upper bound in the integral for $r$ should be in polar coordinates. So far I have the follwing integral: $\sqrt{3} \cdot 4 \int_0^{\pi/2} sin(\theta) d \theta \int_0^{?} r^2 dr$ $\endgroup$ – Kristian Aug 3 '18 at 19:30
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I suggest to use spherical coordinates with

  • $x=r\sin \phi_0 \cos \theta$

  • $y=r\sin \phi_0 \sin \theta$

  • $z=r\cos \phi$

where in that case the angle $\phi_0$ is constant and given by

$$\tan \phi_0 =\frac 1{\sqrt 2} \implies \phi_0=\arctan \frac {\sqrt 2}2 \implies \sin \phi_0=\frac1{\sqrt 3}$$

and the surface element is

$$dS=r\sin \phi_0\,d\theta\, dr=\frac r {\sqrt 3}\,d\theta\, dr$$

Now we need to express $r_{max}(\theta)$ by

  • $z=\sqrt{2(x^2 + y^2)}=\sqrt{2}\,r\sin \phi_0$
  • $z=1+y=1+r\sin \phi_0 \sin \theta$

that is

$$\sqrt{2}\,r\sin \phi_0=1+r\sin \phi_0 \sin \theta \implies r(\sqrt{2}\,\sin \phi_0-\sin \phi_0 \sin \theta)=1 \\\implies r_{max}(\theta)=\frac1{\sin \phi_0(\sqrt{2}\,- \sin \theta)}=\frac{\sqrt 3}{(\sqrt{2}\,- \sin \theta)}$$

therefore we obtain the following set up

$$\int \int_S y dS=\int_0^{2\pi}\, d\theta \int_0^{r_{max}(\theta)} r^2\sin^2 \phi_0 \sin \theta\, dr=\int_0^{2\pi}\, d\theta \int_0^{r_{max}(\theta)} \frac{r^2}3 \sin \theta\, dr$$

By numerical evaluation we obtain $\int \int_S y dS= \sqrt 6 \pi$.

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  • $\begingroup$ I was about to reply almost the same, you beat me to it. I'd like to point out that this is an iterated integral and it can get ugly. $\endgroup$ – Niki Di Giano Aug 3 '18 at 20:20
  • $\begingroup$ @NikiDiGiano I'm sorry! Do you agree with the final set up? $\endgroup$ – gimusi Aug 3 '18 at 20:21
  • $\begingroup$ No problem at all. Although I do find the expression for the ellipse somewhat different, and I used cylindrical coordinates. Namely, all my results are scaled by a factor $\sin(\phi_0)$ with respect to yours, but it could just be due to the change in coordinates. $\endgroup$ – Niki Di Giano Aug 3 '18 at 20:33
  • $\begingroup$ @NikiDiGiano If you show your solution here I can take a look! $\endgroup$ – gimusi Aug 3 '18 at 20:33
  • $\begingroup$ Thanks a lot for your input! Just a brief comment - there must be a small mistake somewhere as the back of my book states that the answer should be $\sqrt{6} \pi$. So the factor of 9 should not be there. $\endgroup$ – Kristian Aug 3 '18 at 20:44
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The domain of integration is the projection on the $xy$-plane of the intersection between the plane and the cone (an ellipse in 3D space). Using cylindrical coordinates we get: $$ x = r\cos \theta\\y=r\sin\theta\\ z =z\\ \sqrt 2 r= r\sin\theta + 1 $$ We can thus determine the expression for the ellipse: $$ r = \frac{1}{\sqrt 2 - \sin \theta} = \rho(\theta)$$ The area element with these coordinates has the form: $$ dS = \sqrt 3 dx dy = \sqrt 3 r drd\theta $$ Which yields the final integral: $$ \sqrt 3 \int_{0}^{2\pi} \sin \theta \int_{0}^{\rho(\theta)} r^2 dr d\theta $$

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  • $\begingroup$ Very nice solution, the final result should be the same. Your ellipse is different since you are representing $r$ on x-y plane whereas mine $r $ is the spherical radius. $\endgroup$ – gimusi Aug 3 '18 at 21:00
  • $\begingroup$ Thanks a lot! This is the approach I tried, but didn't think of setting up the equation $\sqrt{2}r = r\sin(\theta) + 1$ (which I now see is obvious!). $\endgroup$ – Kristian Aug 3 '18 at 21:01
  • $\begingroup$ Thank you @gimusi and happy I could still be helpful. $\endgroup$ – Niki Di Giano Aug 3 '18 at 21:06

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