Computing the area of the intersection of a sphere with a cone

Question

Find the area $A$ of the surface $S$. $S$ is the part of the sphere $x^2 + y^2 + z^2 = 8$ that lies inside the cone $z^2 = x^2 + y^2$.

I know I have to use the double integral surface area formula but I don't know how to set up the integral.

Thanks

Answer 1

You can use sperical coordinates: $$x=\sqrt{8} \cos s \cos t,$$ $$y=\sqrt{8} \cos s \sin t,$$ $$z=\sqrt{8} \sin s,$$ The bounds for $t$ is $0\leq t\leq 2\pi$. To determine the bounds for $s$, note that the surface lies inside the cone $z^2=x^2+y^2$, we have $z^2\geq x^2+y^2$: $$8\sin^2 s\geq 8\cos^2 s\quad\Rightarrow\quad \frac{\pi}{4}\leq s\leq\frac{\pi}{2}.$$ Then you can apply the formula.