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Find the surface area of the portion of the sphere $x^2 + y^2 + z^2 =3c^2$ within the paraboloid $2cz =x^2+y^2$ using spherical coordinates. ($c$ is a positive constant)

I've found it in cartesian coordinates and then polar coordinates by taking limits $r$ from $0$ to $\sqrt 2 c$ and $\theta$ from $0$ to $2\pi$ and got the answer $4\sqrt 3\pi c^3$. But for the spherical coordinates I am getting the limits $\phi$ from $\arccos(\sqrt 3)$ to $\pi/2$ and $\theta$ from $2\pi$ to $0$ and the answer is $6\pi/\sqrt3 c^2$ which is $2c$ times less than in cartesian coordinates. What am I doing wrong?

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The limits for $\phi$ are wrong. Note that $\sqrt{3}>1$ and $\arccos(x)$ is defined in $[-1,1]$. Solving the equation $2cz+z^2=3c^2$ we get $z=c$ and $z=-3c$ (which is not acceptable because $x^2+y^2=2cz=-6c^2<0$). Hence $\phi$ goes from $0$ (north pole) to $\arccos(1/\sqrt{3})$. Therefore $$S=3c^2\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\arccos(1/\sqrt{3})}\sin(\phi) d\phi d\theta=6\pi c^2[-\cos(\phi)]_0^{\arccos(1/\sqrt{3})}=2\pi c^2(3-\sqrt{3}).$$

P.S. Your result $4\sqrt 3\pi c^3$ seems to be a volume not a surface: $$S=\iint_{x^2+y^2\leq 2c^2}\frac{\sqrt{3}c}{\sqrt{3c^2-x^2-y^2}}dx dy= 2\pi\sqrt{3}c\int_{\rho=0}^{\sqrt{2}c}\frac{\rho}{\sqrt{3c^2-\rho^2}}d\rho=2\pi c^2(3-\sqrt{3})$$

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  • $\begingroup$ I used that method to get my final answer.since surface is double integral of root(1+fx^2+fy^2) which becomes root(3c^2)/root(3c^2-r^2). c is a positive constant and r is from 0 to root2*c. So that's the answer I get. Git any ideas?(c is positive so it doesn't depend on angle anymore) $\endgroup$ – Usama Notkani Mar 6 '18 at 10:25
  • $\begingroup$ @UsamaNotkani See my P.S. $\endgroup$ – Robert Z Mar 6 '18 at 10:40
  • $\begingroup$ Ok I get what you're saying thanks. I think I made a mistake in one calculation which I am starting to see now that you've written it out $\endgroup$ – Usama Notkani Mar 7 '18 at 5:57

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