Surface integral in spherical coordinates question

1. May 16, 2009

compliant

1. The problem statement, all variables and given/known data
Find the surface area of the portion of the sphere $$x^2 + y^2 + z^2 = 3c^2$$ within the paraboloid $$2cz = x^2 + y^2$$ using spherical coordinates. (c is a constant)

2. Relevant equations

3. The attempt at a solution
I converted all the x's to $$\rho sin\phi cos\theta$$, y's to $$\rho sin\phi sin\theta$$, z's to $$\rho cos\theta$$. Thus, the vector form of the sphere is $$\vec{r}(\varphi,\theta) = {\rho sin\phi cos\theta}\hat{i} + {\rho sin\phi sin\theta}\hat{j} + {\rho cos\theta}\hat{k}$$. Thus, $${\left\|{{\vec{r}}_{\varphi}}\times {{\vec{r}}_{\theta}}\right\|} = {\sqrt{3c^2}} sin \varphi$$

Now I want to find the bounds of integration for the surface integral. I know theta goes from 0 to $$2\pi$$, but for $$\varphi$$, I get this mess while trying to equate the paraboloid to the sphere:

$$2c({\rho cos\theta}) = {{\rho}^2 sin^2 \varphi}$$

When $$\theta = 0, \varphi = sin^{-1}({\sqrt{2c/{\rho}}})$$, and when $$\theta = 2\pi, \varphi = sin^{-1}({\sqrt{2c/{\rho}}})$$.

??????

I really hate vector calculus.

2. May 16, 2009

AUMathTutor

I would first identify the region using whatever coordinates I wanted to. Then you can carry out the integration using spherical coordinates. Equating them in Cartesian coordinates provides some very useful information.

How do you go about this? Well, you know that wherever the paraboloid and the sphere intersect, the circular cross sections in the x-y parallel plane will be of the same size. Ergo, where they meet, the quantity x^2 + y^2 will be the same. Rearranging and equating, we arrive at

3c^2 - z^2 = 2cz

Rearranging,

z^2 + 2cz - 3c^2 = 0

I think z comes out to be: [-2c +- sqrt(4c^2 + 12c^2)]/2 = -3c OR +c.

So you're right that theta goes from 0 to 2PI; the only question now is what the phi angle should do. You know it has to go from the top of the sphere down until the point where z = c (assuming c positive, this is the only sensible solution). So when is z = c? well, you have a circle of radius c*sqrt(3), and you need to be c up on it... so you know that c*sqrt(3) * cos(phi) = c.

I might have made some basic mistakes somewhere, but the idea is sound. Does that make sense? I wouldn't change things to spherical coordinates until you literally begin doing the integration. It's just so much easier to work with Cartesian as long as possible.

3. May 16, 2009

compliant

Well, the other part of the question (which I had done already) was doing it in Cartesians, so I know the two surfaces intersect at z = c, as you stated.

So,
$$(c \sqrt{3}) cos(\varphi) = c$$
$$cos (\varphi) = 1/{\sqrt{3}}$$
$$\varphi =$$ a really ugly number.

Should I just keep this as $$cos^{-1} (1/{\sqrt{3}})$$ ? I guess it wouldn't be a bad idea. So then my integration is

$$\int_{0}^{2\pi} \int_{0}^{cos^{-1} (1/{\sqrt{3}})} {3c^2 sin(\varphi)} d\varphi d\theta$$
$$= {-6\pi c^2 cos(\varphi)]^{cos^{-1} (1/{\sqrt{3}})}_{0}}$$
$$= -6\pi c^2({1/{\sqrt{3}}}-1)$$
$$= 6\pi c^2(1-{1/{\sqrt{3}}})$$

Last edited: May 16, 2009
4. May 16, 2009

compliant

But now I have a new problem with the surface area done in Cartesians. When z=c,

$$2c^2 = x^2 + y^2$$

where
$$-\sqrt{2c^2 - x^2} \leq y \leq \sqrt{2c^2 - x^2}$$

and
$$-\sqrt{2c^2} \leq x \leq \sqrt{2c^2}$$

dS for the sphere is $${\sqrt{\frac{3c^2}{3c^2 - x^2 - y^2}}}dy dx$$

Surface area = $$\int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} \int_{-\sqrt{2c^2 - x^2}}^{\sqrt{2c^2 - x^2}} {{\sqrt{\frac{3c^2}{3c^2 - x^2 - y^2}}}dy dx}$$
= $$c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} \int_{-\sqrt{2c^2 - x^2}}^{\sqrt{2c^2 - x^2}} ({{\sqrt{3c^2 - x^2 - y^2}})^{-{\frac{1}{2}}}}dy dx}$$
= $$-c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} ({\sqrt{3c^2 - x^2 - y^2}})^{{\frac{1}{2}}}} ]^{\sqrt{2c^2 - x^2}}_{-\sqrt{2c^2 - x^2}} dx}$$
= $$-c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} (c-c) dx$$

But then....the integrand becomes zero. I don't think that's supposed to happen. =/

Last edited: May 16, 2009
5. May 16, 2009

HallsofIvy

The sphere and paraboloid intersect when $$2cz+ z^2= 3c^2$$. Completing the square, $$z^2+ 2cz+ c^2= 4c^2$$ or $$z+ c= \pm 2c$$, z= c or z= -3c.

If c is positive, since $$2cz= x^2+ y^2$$, z is positive: we must have z= c so $$x^2+ y^2= 2c^2$$, a circle or radius $$\sqrt{2}c$$.

If c is negative, z is negative: we must have z= -3c so $$x^2+y^2+ 9c^2= 3c^2$$ or $$x^2+ y^2= -6c^2$$. That's impossible so c must be positive and the two surfaces intersect on the circle $$x^2+ y^2= 2c^2$$, z= c.

Obviously $$\theta$$ ranges from 0 to $$2\pi$$. To find the limits on $$\phi$$, not that a point on the boundary circle is distance $$\sqrt{2}c$$ from the z-axis and at height $$z= c$$. $$\phi$$ is given by $$tan(\phi)= \sqrt{2}c/c= \sqrt{2}$$. $$\phi$$ ranges from 0 to $$tan^{-1}(\sqrt{2})$$.

Because you are asked specifically to find the surface area, you should use spherical coordinates, with $\rho= \sqrt{2}c$ to get parametric equations for the surface:
$$\vec{r}(\theta,\phi)= \sqrt{2}c cos(\theta)sin(\phi)\vec{i}+ \sqrt{2}c sin(\theta)sin(\phi)\vec{j}+ \sqrt{2}c cos(\phi)\vec{j}$$

Last edited by a moderator: May 16, 2009
6. May 16, 2009

compliant

Yes, but unforunately, my Cartesian answer not the same.

I see now why though:

$$c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} \int_{-\sqrt{2c^2 - x^2}}^{\sqrt{2c^2 - x^2}} ({{\sqrt{3c^2 - x^2 - y^2}})^{-{\frac{1}{2}}}}dy dx}$$
= $$-c\sqrt{3} \int_{-\sqrt{2c^2}}^{\sqrt{2c^2}} ({\sqrt{3c^2 - x^2 - y^2}})^{{\frac{1}{2}}}} ]^{\sqrt{2c^2 - x^2}}_{-\sqrt{2c^2 - x^2}} dx}$$

Obviously this isn't what one should get after integrating that term. And I can't exactly stick in a 1/y right in front of the bracket to cancel out the y term that comes out after doing the Chain Rule. Does anyone know how to integrate it? =/