parametrization of surfaces and area

Question

The question is:

Paraboloid $z=x^2+y^2 $ divides the sphere $x^2+y^2+z^2=1$ into two parts, calculate the area of each of these surfaces.

I know that i need to use $\iint |n| dS$ but i need to first find the normal to this surfaces and for that i need to parameterize each part, i do not know much about parametrization of part of surfaces, how can i use $\theta,\phi$ for this parametrization or should i use something else?

Any suggestion would be great, thanks

Answer 1

The intersection of the paraboloid and the sphere will be a circle, which occurs at

$$z+z^2=1 \implies z = \frac{\sqrt{5}-1}{2} \equiv z_0$$

From here there are many avenues you can take to calculate the area of the top cap (the bottom one is just $4\pi$ minus this area).

$\textbf{Option 1}$: Directly

Parametrize with spherical coordinates, $r = 1$ and $J = \frac{\frac{\partial(x,y,z)}{\partial r}}{\left|\frac{\partial(x,y,z)}{\partial r}\right|}\Biggr|_{r=1} = \sin\theta$

$$\iint\limits_{x^2+y^2+z^2=1\cap z\geq z_0} dS = \int_0^{2\pi}\int_0^{\cos^{-1}(z_0)}\sin\theta\;d\theta d\phi = 2\pi(1-z_0) = (3-\sqrt{5})\pi$$

$\textbf{Option 2}$: Surface area formula

Plug in $z = \sqrt{1-x^2-y^2}$

$$\iint_{x^2+y^2\leq z_0} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA = \int_0^{2\pi}\int_0^{\sqrt{z_0}} \frac{r}{\sqrt{1-r^2}}drd\theta = 2\pi(1-\sqrt{1-z_0}) = 2\pi(1-z_0)$$

since $z_0^2 = 1-z_0$

$\textbf{Option 3}$: Divergence Theorem

Consider the vector field $F(x,y,z) = \left(0,0,\frac{1}{z}\right)$ and close the top spherical cap with the plane $z=z_0$. Using $n = (x,y,z)$ for the spherical cap, we have that

$$\iint\limits_{\text{cap}}dS = \iint_{x^2+y^2+z^2=1\cap z\geq z_0}\left(0,0,\frac{1}{z}\right)\cdot(x,y,z)dS = \iiint\limits_{x^2+y^2+z^2\leq1 \cap z\geq z_0}-\frac{1}{z^2}dV + \iint_{x^2+y^2+z^2\leq 1 \cap z=z_0}\frac{1}{z_0}dS$$

$$= \int_0^{2\pi}\int_{z_0}^1\int_0^{\cos^{-1}\left(\frac{z_0}{r}\right)}-\frac{\sin\theta}{\cos^2\theta}\:d\theta dr d\phi + \frac{1}{z_0}\iint_{x^2+y^2\leq z_0} dA = 2\pi \int_{z_0}^1 1 - \frac{r}{z_0}\:dr+\pi$$

which will simplify to the other options.

Answer 2

If you are using parametrization,

$f(x,y,z) = (x, y, g(x,y))$ where $g(x,y) = \sqrt{1-x^2-y^2}$

You can check that $|f_x \times f_y| = \sqrt{{1 + (\frac{\partial g}{\partial x}})^2 + (\frac{\partial g}{\partial y})^2} = \frac{1}{\sqrt{1-x^2-y^2}}$

Now the intersection of both surfaces is given by $z^2 = 1 - x^2 - y^2 = 1 - z$

$\implies z = \frac{\sqrt5 - 1}{2}$ (you have a negative value of $z$ from the quadratic that you ignore given the surfaces intersect above $z = 0$).

So you are now look at projection in $XY-$plane of a circle given by $x^2+y^2 = \frac{\sqrt5 - 1}{2}$.

So your integral becomes $\displaystyle \int_0^{2\pi} \int_0^{a}\frac{r}{\sqrt{1-r^2}} \ dr \ d\theta$

where $a = \sqrt{\frac{\sqrt5 - 1}{2}}$

If you are parametrizing in spherical coordinates,

$r(\theta, \phi) = (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$

Again you can check that $|r_{\phi} \times r_{\theta}| = \sin \phi$

Also you can check that paraboloid surface is given by $\rho = \frac{\cos \phi}{\sin^2\phi}$

At intersection of both surfaces, $\frac{\cos \phi}{\sin^2\phi} = 1 \implies \cos^2\phi+\cos\phi-1= 0$

$\implies \phi = \arccos(\frac{\sqrt5 - 1}{2})$

So the integral becomes

$\displaystyle \int_0^{2\pi} \int_0^{\arccos(\frac{\sqrt5 - 1}{2})} \sin\phi \ d\phi \ d\theta$