Show that the surface of half hemisphere

Question

Show that the surface of half hemisphere is equal to

$2 \pi r^2$

by using surface integral.

Answer 1

Using spherical coordinates $(r,\theta, \varphi)$ and fixing $r=R,$ we have

$$A = \iint_\Omega dS$$ where our surface element for a given radius is given by $dS = R^2 \sin(\theta)d\theta d\varphi \; .$ This means we have

$$A = \int_0^\pi \int_0^\pi R^2 \sin(\theta)d\theta d\varphi = 2\pi R^2$$ for the half-sphere.

Answer 2

We can write the hemisphere with radius r, above the xy-plane, as $x= rcos(\theta)sin(\phi)$, $y= rsin(\theta)sin(\phi)$, $z=r cos(\phi)$ with $\theta$ from 0 to $2\pi$,$\phi$ from 0 to $\frac{\pi}{2}$.

The "differential of surface area" for a sphere or radius r in those coordinates is $r^2 sin(\phi)d\theta d\phi$ so the surface area of hemisphere is given by $\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2} r^2 sin(\phi) d\theta d\phi= r^2\left(\int_0^{2\pi} d\theta\right)\left(\int_0^{\pi/2} sin(\phi)d\phi\right)$ $= r^2 (2\pi)\left(-cos(\phi)\right)_0^{\pi/2}= 2\pi r^1\left(0- (-1)\right)= 2\pi r^2$.