Help finding parameterization for a surface integral

Question

Find the surface area above the xy plane which is defined by: $x^2+y^2+z^2=1, x^2+y^2 \leq x$

I am first trying to find a parameterization which should be simple enough to work with.

I though about using spherical coordinates: $$ x=\sin(\theta)\cos(\phi)$$ $$ y=\sin(\theta)\sin(\phi) $$ $$z=\sqrt{1+\sin^2(\theta)}$$

Which should work but the problem here is that I'm not sure how to express where $\phi$ is defined.

Because we have $0 \leq \theta \leq \frac{\pi}{2}$, and $\sin^2(\theta) \leq \sin(\theta) \cos(\phi) \Rightarrow \sin(\theta) \leq \cos(\phi) $.

Help appreciated.

Answer 1

Note that$$x^2+y^2\leqslant x\iff\left(x-\frac12\right)^2+y^2\leqslant\frac14.$$So, you're after the area of the surface $z=\sqrt{1-x^2-y^2}$ with $x$ and $y$ such that $\left(x-\frac12\right)^2+y^2\leqslant\frac14$. This area is equal to\begin{multline}\int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}\sqrt{\frac{x^2}{1-x^2-y^2}+\frac{y^2}{1-x^2-y^2}+1}\,\mathrm dy\,\mathrm dx=\\=\int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}\frac1{\sqrt{1-x^2-y^2}}\,\mathrm dy\,\mathrm dx.\end{multline}In polar coordinates, this is\begin{align}\int_{-\pi/2}^{\pi/2}\int_0^{\cos(\theta)}\frac\rho{\sqrt{1-\rho^2}}\,\mathrm d\rho\,\mathrm d\theta&=\int_{-\pi/2}^{\pi/2}\left[-\sqrt{1-\rho^2}\right]_{\rho=0}^{\rho=\cos(\theta)}\,\mathrm d\theta\\&=\int_{-\pi/2}^{\pi/2}1-\bigl|\sin(\theta)\bigr|\,\mathrm d\theta\\&=2\int_0^{\pi/2}1-\sin(\theta)\,\mathrm d\theta\\&=\pi-2.\end{align}

Answer 2

Unit sphere: $x^2+y^2+z^2=1$, Cylinder: $x^2+y^2 \leq x, z \geq 0$. Please note the equality in equation of the sphere and inequality in the equation of the cylinder which means we need to find the surface area of the sphere that is bound by the cylinder for $z \geq 0$.

As you mentioned in spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$.

For sphere, the surface is $\rho = 1$

Equation of the cylinder $x^2+y^2 \leq x$ in spherical coordinates becomes
$\rho = \cos \theta \csc \phi$

At the intersection of the cylinder and the sphere,

$\rho = \cos\theta \csc \phi = 1 \implies \phi = \arcsin(\cos\theta)$

Also note the cylinder has radius of $\frac{1}{2}$ and its center is at $(\frac{1}{2}, 0, z)$. For the cylinder in $XY$ plane, $\theta$ varies from $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

For the given sphere, $|r'_{\theta} \times r'_{\phi}| = \sin\phi$

So the integral for the surface area is

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{\arcsin(\cos\theta)} \sin \phi \ d\phi \ d\theta \approx 1.1416$

In cylindrical coordinates,

$x = \rho \cos\theta, y = \rho \sin\theta, z$

Sphere is given by $r^2 + z^2 = 1$. Cylinder is given by $r = \cos\theta, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

For the given sphere, $|r'_{\rho} \times r'_{\theta}| = \frac{\rho}{\sqrt{1-\rho^2}}$

So the surface area is

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{\cos\theta} \frac{\rho}{\sqrt{1-\rho^2}} \ d\rho \ d\theta \approx 1.1416$