Find the range of surface integral using spherical coordinates

Question

Let $S$ be a section of sphere $x^2+y^2+z^2=3$ with $x\ge1$ and $y\ge1$(like wedge shape). Compute the area of $S$ finding the range of surface integral over $S$ via spherical coordinates.

If the sphere $x^2+y^2+z^2=3$ parametrized $X(\theta,\phi)=(\sqrt{3}\cos\theta\sin\phi,\sqrt{3}\sin\theta\sin\phi,\sqrt{3}\cos\phi)$ with $0\le\theta\le2\pi,~0\le\phi\le\pi$, then how to find the range of $\theta$ for $S$?

I tried $x=\sqrt{3}\cos\theta\sin\phi\ge1$ and $y=\sqrt{3}\sin\theta\sin\phi\ge1$, but i couldn't find the common range for $\theta$.

Also, is it true that the range of $\phi$ is $\cos^{-1}(1/\sqrt{3})\le\phi\le\cos^{-1}(-1/\sqrt{3})$?

I'm not sure about that. Give some advice. Thank you!

Answer 1

HINT:

Actually Cartesian coordinates are more suitable for the computation, but problem is given that way;

To workaround, since

$$ r=R \cos \phi , R= \sqrt{3},$$

$$ x_1= R \cos \phi \cos\theta ;\, y_1= R \cos \phi \sin \theta $$ are constants and the intersection point is $$ (x,y,z)= (1,1,\pm 1), $$

it helps in finding integration limits.

Answer 2

I solve the problem as i wished:

$$\iint_{S_{0}}dS=\int_{\cos^{-1}\left(\tfrac{1}{\sqrt{3}}\right)}^{\pi-\cos^{-1}\left(\tfrac{1}{\sqrt{3}}\right)}\int_{\sin^{-1}\left(\tfrac{1}{\sqrt{3}\sin\phi}\right)}^{\cos^{-1}\left(\tfrac{1}{\sqrt{3}\sin\phi}\right)}3\sin\phi\,d\theta\,d\phi=(2-\sqrt{3})\pi.$$