Double integratration to find surface area of a cut out sphere

Question

How to solve the surface area of a sphere $x^2 + y^2 + z^2 = r^2$ cut out by $x^2 +y^2 = r^2 - (r \cos\theta)^2$ that is above $xy$-plane

$$\frac{dz}{dx} = \frac{-x}{\sqrt{r^2-x^2-y^2}}$$

$$\frac{dz}{dy} = \frac{-y}{\sqrt{r^2-x^2-y^2}}$$

$$\begin{align} Serface Area & = \iint \sqrt{1+(\frac{dz}{dx})^2 +(\frac{dz}{dy})^2} dx\,dy\\& = \iint \sqrt{1+(\frac{-x}{\sqrt{r^2-x^2-y^2}})^2 + (\frac{-y}{\sqrt{r^2-x^2-y^2}})^2 }dx\,dy\\ &= \iint \sqrt{\frac{r^2}{r^2-x^2-y^2}} dx\,dy \\ &= \int^{2\pi}_{0}\int^{\sqrt{r^2 - (r \cos\theta)^2}}_{0} \sqrt{\frac{r^2}{r^2-(r^2cos\theta+r^2sin\theta))}}r\,dr\,d\theta \\ & = \int^{2\pi}_{0}\int^{\sqrt{r^2 - (r \cos\theta)^2}}_{0} \sqrt{\frac{r^2}{r^2-r^2}}r\,dr\,d\theta \end{align}$$

After changing it to polar coordinate with $x=rcos\theta$ and $y=rsin\theta$, the denominator is $0$. What am I doing wrong here?

Answer 1

As to why you are having trouble with the double integral, in the work you have shown, there is some confusion regarding polar coordinates. $r$ and $\theta$ are already taken up in the specification of the problem. As I mention in the other answer, the plane is at $z=r \cos \theta$. So, if you want to parametrize the plane for the carrying out a double integral, I suggest another set of variables for that purpose, say $\rho$ and $\varphi$. As you have noted correctly, the double integral for the surface area is

$$S=\int \int \frac{r}{\sqrt{r^2-x^2-y^2}}dxdy $$

Making the change of variables $x=\rho \cos \theta$, $y=\rho \sin \theta$

$$S=\int_0^{2 \pi} \int_0^{r \sin \theta} \frac{r}{\sqrt{r^2-\rho^2}}\rho d\rho \varphi = 2 \pi r \int_0^{r \sin \theta} \frac{1}{\sqrt{r^2-\rho^2}}\rho d\rho$$

We can integrate that easily with a change of variables $t^2=r^2-\rho^2$ to get $$S=2\pi r^2(1- \cos\theta)$$

Answer 2

You can do this without using a double integral. The question is basically asking for the surface area of the sphere above the plane $z=h$ where $h=r \cos \theta$. Note that this is a surface area of the surface of revolution of a portion of the circle $x^2+z^2=r^2$ about the z-axis. You may already know how to do that, if not, here is the formula (the page uses $x$ and $y$ instead of your $x$ and $z$, so you need to change accordingly).

Your integral is $$2\pi \int_h^r x \sqrt{1+\left(\frac{dx}{dz}\right)^2} dz$$ where you can get $\frac{dx}{dz}$ by implicitly differentiating $x^2+z^2=r^2$ so that $\frac{dx}{dz} = -\frac{z}{x}$

The surface area is $2 \pi r (r-h) = 2 \pi r^2(1- \cos \theta)$ which is a very neat result that the surface area is simply proportional to height $r-h$ of the cap that the plane is cutting off.