Area of a hemisphere with boundary

Question

This is a question from Creighton Buck's Advanced Calculus book. Question goes as follows:

Find the area of the portion of the upper hemisphere of the sphere with center $(0,0,0)$ and radius $R$ that obeys $x^2+y^2-Ry\le0$.

So far, i have tried to draw it considering the boundary given is a circle. Solving that way gives the wrong answer. Then i tried to write a transformation as:

$T(x,y,z)=(R\cos u\sin v, R\sin u\sin v, R\cos v)$. Then I calculated the area using normals but, either way i get the wrong answer.

What am I missing?

Answer 1

We are told to compute the area of the part $S$ of said hemisphere $H$ which is enclosed in the cylinder $$x^2+\left(y-{R\over2}\right)^2\leq{R^2\over4}\ .$$

Denote by $\theta$ the geographical latitude ($\theta={\pi\over2}$ at the north pole) on $H$. The surface element engendered by an infinitesimal latitude strip of length $\ell$ and width $R\,d\theta$ is then given by $${\rm d}\omega=\ell\cdot R\,d\theta\ .$$

The read arc in the above figure is on the circle of latitude $\theta$ and has radius $r=R\cos\theta$. Note that $$\cos\alpha={r\over R}=\cos\theta\ ,$$ hence $\alpha=\theta$. The length $\ell$ of the red arc therefore is $\ell=2\theta r=2 R\theta\cos\theta$. It folllows that the area in question is given by $${\rm area}(S)=R^2\int_0^{\pi/2}2\theta\>\cos\theta\>d\theta=(\pi-2)R^2\ .$$

Answer 2

I'll try to use the variables you've chosen. They are spherical coordinates but with $v=\theta$ and $u=\phi$ instead of the usual names. The sphere in this coordinate system is simply $r=R$

Limits of integration:

$\phi$ varies between $0$ and $\pi$. $\theta$ varies between $0$ and some value depending on $\phi$. Fixing $\phi=\phi_0$ we can find out the value for $\theta_0$. We have:

for points in the cylinder with coordinates $x_0$ and $y_0$

$\sqrt{x_0^2+y_0^2}\sin\phi=y_0$ and $x_0^2+y_0^2=Ry_0\implies y_0=R\sin^2\phi_0$

and

$R\sin\theta_0=\sqrt{x_0^2+y_0^2}$

So, $R\sin\theta_0=R\sin\phi_0\implies \begin{cases} \theta_0=\phi_0\;,\phi_0\lt\pi/2\\ \theta_0=\pi-\phi_0\;,\pi/2\le\phi_0\le\pi \end{cases}$

By symmetry, the integral between $\phi=\pi/2$ and $\phi=\pi$ is the same as between $\phi=0$ and $\phi=\pi/2$, so (with $R^2\sin\theta d\theta d\phi$ as surface element),

$$A=2\int_{0}^{\pi/2}\int_0^{\phi}R^2\sin\theta d\theta d\phi=$$

$$=2R^2\int_{0}^{\pi/2}(1-\cos\phi)d\phi=R^2(\pi-2)$$