Surface integral - cone below plane 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!
 A: 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 $\int \int_S y dS= \sqrt 6 \pi$.
A: 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:
$$ \sqrt 3 \int_{0}^{2\pi} \sin \theta \int_{0}^{\rho(\theta)} r^2 dr d\theta $$
