# Surface integral using spherical coordinates

I have some troubles with the following surface integral:
The lamp-shade L has the form $$x^2+y^2+z^2=2, z \leq 1$$, decide the energy flow through L from a point-shaped lightsource in origin with intensity $$u(r)=cr/r^3 (W/m^2)$$.

I use the parametrization with spherical coordinates $$r=\sqrt{2} \big(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta \big)$$ where $$\pi/4 \leq \theta \leq \pi, 0 \leq \phi \leq 2\pi$$.

Calculating the unitnormal vector and vectorproduct, I obtain

$$r_{\theta}\prime \times r_{\phi}\prime = 2 \sin \theta \big(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta \big)$$

$$\big| r_{\theta}\prime \times r_{\phi}\prime \big|= 2 \sin \theta$$

Hence, $$N = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$.

My troubles are now with the surface integral and I'm a bit unsure what means with "point-shaped light source in origo"

$$\iint_L u \cdot N dS = \iint_L c \frac{r}{r^3} \cdot (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) dS$$

$$= \iint_L c \frac{1}{r^2} 2 \sin \theta \big(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta \big) d\theta d\phi = ?$$

The solution should equal $$2 \pi c (1+1/\sqrt{2})$$.

Any suggestions is highly appreciated, since I'm not sure what I missed. I assume I should plug in r in the $$u(r)$$ function but the fractions obtained is a mess.
Sincerely, TS

It seems that part of your difficulty is notation. You seem to be using the symbol $$r$$ for two different things: a position vector, and a distance.

Are you sure the original problem is not written as follows?

$$\mathbf u(\mathbf r) = \frac{c\mathbf r}{r^3}\ (\mathrm W/\mathrm m^2).$$

Here the symbol $$\mathbf r$$ is a vector and $$r$$ is a scalar representing the magnitude of $$\mathbf r$$. The difference in the typeface is subtle for this letter but can be distinguished if you look for it.

It follows then that on the surface of the sphere, \begin{align} \mathbf r &=\sqrt{2} (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta),\\ r &= \sqrt 2, \\ \mathbf N &= (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta). \end{align} Therefore $$\mathbf r = r \mathbf N$$.

If the problem is posed like this, you don't have to interpret "point-shaped light source in origo" because the energy flow generated by that source is completely described by the vector function $$\mathbf u(\mathbf r)$$.

In the integral, remember that you are looking at a dot product $$\mathbf u \cdot \mathbf N$$, so you need to identify the vectors on both sides and evaluate the dot product accordingly. It's not completely clear how you ended up with the factor $$2\sin\theta$$ in your last integral; I can only guess it came from the expression for $$r_{\theta}' \times r_{\phi}'$$. That expression did all the work you can expect from it when you used it to calculate $$\mathbf N$$; it has nothing more to do with the dot product $$\mathbf u \cdot \mathbf N$$.

As a hint for a way to save a lot of effort calculating the dot product, what is the dot product of a unit vector with itself?

• You are completely right, thank you very much sir. Have a great day. Commented Aug 1 at 14:28

A "point-shaped lightsource" (or simply, point source) is just like what it sounds: a power source concentrated at a point. The intensity follows an inverse square law: $$|\vec{u}| = \frac{c}{r^2}$$, where $$4 \pi c$$ is the power leaving the source. The direction of the flux is radially outward from the source.

So, you really don't even need integration, as the source is at the center of your lamp, so the flux from it is everywhere normal to your (spherical) surface. The flux intensity at distance $$\sqrt{2}$$ from the source is $$\frac{1}{2} c$$, and the area of the lamp $$L$$ being illuminated is $$A = (\sqrt{2})^2 \Omega$$, where $$\Omega = 2 \pi \left( 1 + \frac{\sqrt{2}}{2} \right) = \pi (2 + \sqrt{2})$$ is the enclosed solid angle. Therefore, the integrated flux is

$$\left( \frac{1}{2} c \right) (2 \pi) (2 + \sqrt{2}) = \pi c (2 + \sqrt{2})$$

It's not nessecary to calculate so much: The source is pointlike, the intensity falls of by $$W\ \frac{1}{4 \pi r ^2}$$ over any sphere of radius $$r$$ by constancy of the energy transport per second. The flow is normal to the sphere. It follows, that an area $$A$$ has a total flow of $$W \ \frac{A}{4 \pi r ^2}$$.

Now use spherical coordinates, but replace $$z\ = \ \cos \theta$$ $$X=r \ \left(\cos \phi \ \sqrt{1-z^2}, \sin \phi \ \sqrt{1-z^2}, z \right)$$

The volume in these coordinates by orthogonality is simply the metric product of lenghts

$$d\text {Vol} \ = dr \wedge r d\theta \wedge r \sin \theta d\phi = r^2\ dr \wedge dz \wedge d\phi.$$ For functions at constant r independent of $$\phi$$ we have the area

$$d A = 2 \ \pi\ r^2\ dz$$ in other words, the area between two circles of latitude is proportional to their hight difference along the z-axis.

This yields total area with $$r=2, -2< 2 z < 1$$

$$W \ \frac{r^2}{4 \pi \ r^2}\int_{-1}^{\frac{1}{2}} dz \ \ \int_0^{2\pi} d\phi = \frac{3}{4} \ W$$