Calc 3 : "Find the surface area of the part of the sphere $x^2+y^2+z^2= r^2$ where $2r/3<z$

Question

Ive been struggling with this problem. I understand that as a surface integral problem I need to parameterize in terms of spherical coordinates. What I don't know is how does $2r/3<z$ affect the bounds and how I am supposed to integrate this. Any tips would be greatly appreciated. Thanks!

Answer 1

We can use spherical coordinates with

• $x=r\sin \phi \cos \theta$
• $y=r\sin \phi \sin \theta$
• $z=r\cos \phi$

with

• $dS=r^2\sin \phi \,d\phi\,d\theta$

where

• $r$ is constant
• $\theta \in (0,2\pi)$
• $\phi \in (0,\phi_0)$ with $\phi_0=\arccos\left(\frac23 \right)$

therefore

$$S=\int_0^{2\pi} \, d\theta \int_0^{\phi_0} r^2\sin \phi \,d\phi=2\pi r^2\left[-\cos \phi\right]_0^{\phi_0}=-\frac{4}3\pi r^2+2\pi r^2=\frac23 \pi r^2$$

As an alternative and as a check recall taht the area of a spherical sector is given by

• $S=2\pi r h\quad h=r-\frac23 r=\frac13 r\implies S=\frac23 \pi r^2$