# Calculating the area of a surface given by a set $S$

$$S=\{(x,y,z):x^2+y^2+z^2=4, (x-1)^2+y^2 \leq 1 \}$$.

$$x^2+y^2+z^2=4 \iff \frac{x^2}{4}+\frac{y^2}{4} +\frac{z^2}{4}=1$$

I'm not exactly sure what to parametrize the set $$S$$ by I thought of using spherical coordinates, in particular $$G(\theta, \phi)=(\frac{1}{2}cos\theta sin\phi, \frac{1}{2}sin\theta sin\phi, \frac{1}{2}cos\phi)$$

$$(x-1)^2+y^2 \leq 1$$ I'm not sure how this is contributing to the work..

• Think about what $S$ looks like. The first equation gives a sphere of radius 4 centered at the origin and the second equation restricts to the portion of the sphere inside the closed right circular cylinder centered at $(1,0)$ of radius $1$. As such, I think cylindrical coordinates would probably be helpful. Admittedly, I haven't actually attempted any computations. – Gary Moon Mar 15 at 2:19

The set here is a pair of spherical caps - the result of intersecting a sphere with the inside of an off-center cylinder. The projection of the sphere in the $$xy$$-plane is a circle of radius $$2$$, while the projection of the cylinder is a circle of radius $$1$$, off-center by $$1$$. They touch on the boundary, but the smaller circle is entirely contained in the larger one - which means we get a cap with $$z>0$$ and another with $$z<0$$.

So, then, parametrizing. First, rectangular coordinates:
We have $$(x-1)^2+y^2\le 1$$, and $$z=\pm\sqrt{4-x^2-y^2}$$. That gives us two parametrized regions: $$0\le x\le 2, -\sqrt{1-(x-1)^2}\le y\le\sqrt{1-(x-1)^2}, z(x,y)=\sqrt{4-x^2-y^2}$$ $$0\le x\le 2, -\sqrt{1-(x-1)^2}\le y\le\sqrt{1-(x-1)^2}, z(x,y)=-\sqrt{4-x^2-y^2}$$

Well, that's a lot of square roots. Definitely not the easiest to work with.

How about cylindrical coordinates? We've got two circles here, so there are two ways to go. First, cylindrical with respect to the small circle: $$0\le r\le 1, 0\le\theta\le 2\pi, x = 1+r\cos\theta, y = r\sin\theta, z=\sqrt{4-(1+r\cos\theta)^2-r^2\sin^2\theta}$$ $$0\le r\le 1, 0\le\theta\le 2\pi, x = 1+r\cos\theta, y = r\sin\theta, z=-\sqrt{4-(1+r\cos\theta)^2-r^2\sin^2\theta}$$ Those $$z$$ formulas simplify a bit, but there will still be a $$\cos\theta$$ term in there.

Next, cylindrical with respect to the large circle: $$(r\cos\theta-1)^2+(r\sin\theta)^2\le 1,x=r\cos\theta,y=r\sin\theta,z=\sqrt{4-r^2}$$ $$(r\cos\theta-1)^2+(r\sin\theta)^2\le 1,x=r\cos\theta,y=r\sin\theta,z=-\sqrt{4-r^2}$$ In this option, we've shunted the more complicated parts off into the bounds for $$r$$ and $$\theta$$. Expanding that inequality out, it becomes \begin{align*}r^2\cos^2\theta-2r\cos\theta+1+r^2\sin^2\theta &\le 1\\ r^2 &\le 2r\cos\theta\end{align*} So then, $$-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$$ and $$0\le r\le 2\cos\theta$$. That looks like the way to go.

So, now that you have a parametrization, can you set up and solve the area integral?

• Thank you very much for your thorough answer. Do you have any general advice for parametrizing a set by any chance? I just can't seem to understand or see how to do it... like everything else I am good with but these parametrization is what gives me the most amount of trouble.. Do you have any recommended book where it goes through plenty of these examples that I can from? – javacoder Mar 15 at 2:52
• Honestly, I don't have any formal process here. It's a matter of building things up from a library of standard elements - sometimes writing one thing as a function of others, sometimes using angles to handle a circle or disk, sometimes basing parameters on a process used to generate the set. Doing examples is definitely a good way to learn, but I don't know any good sources for them. – jmerry Mar 15 at 3:05

Using spherical coordinates could be a better option. Use $$x = rcos\theta cos\phi$$ $$y = rsin\theta cos\phi$$ $$z = rsin\phi$$ Where in this case $$r = 2$$

Then the second equation becomes $$r^2cos^2\phi - 2rcos\theta cos\phi = 0$$ $$\implies rcos\phi = 2cos\theta$$ $$\implies cos\phi = cos\theta$$ $$\implies \phi = \theta$$

The integral for the surface is

$$2\int_0^\frac{\pi}{2}\int_0^{\theta }r^2cos\phi d\phi d\theta$$ $$= 8\int_0^\frac{\pi}{2}sin\theta d\theta = 8$$