# Surface integral of a vector function. Spherical coordinates

First I'm sorry for asking this kind of a question. I know there are a lot explanations online but I don't understand them because this is something new for me.

I'm studying engineering. On my first year I had math but we never learned anything about surface integrals of vector function. And now, on second year, I have a problem. I need to calculate surface integral of vector function (current density through a sphere cap) using spherical coordinates. I looked online but nothing was helpful.

I don't know how to calculate Jacobian and how to express dS using spherical coordinates.

Is there a theorem I can use? Or can you link me to a good website about integration of vector functions?

I have vector function:

$\mathbf J=-2.4 \mathbf a_z$

I need to calculate itegral:

$$I=\int_S\mathbf Jd\mathbf S$$ (Current through spherical cap)

$r=5mm$

$\theta \in [0,\frac{\pi}{2}]$

$\phi\in [0,2\pi]$

I have solution that I don't quite understand and I'm not sure if it's correct:

$d\mathbf S=r^2sin\theta d\phi d\theta \mathbf a_r$;

$\mathbf a_z*\mathbf a_r=cos\theta$;

$d(sin\theta)=cos\theta d\theta$

Again, I'm sorry for this kind of a question but this site was so helpful before that's why I'm asking this here.

• People can't tell you how to calculate integrals they can't see. – Professor Vector Dec 17 '17 at 18:20
• I'm sorry I was going to write it but I forgot. Thank you – Lamija37 Dec 17 '17 at 18:24
• When I was in doubt in vector calculus I turned to Paul's Online notes, I found it to be reasonably nice in explaining things I was getting wrong.tutorial.math.lamar.edu . Plus they aren't just about calc 3, but include other topics as well for the aspiring STEM student – Triatticus Dec 17 '17 at 18:55
• What do you mean by $a_z$ ? You should be able to substitute $dS$ in and integrate over $\phi$ and $\theta$ – Christian Fieldhouse Dec 17 '17 at 19:14
• @Lamija37 write down $(x,y,z)$ in terms of polar co-ordinates, and take the matrix of partial derivative with respect to $\theta, \phi, r$ and take the determinant. This will give you $dS$. – Andres Mejia Dec 17 '17 at 19:27

The Jacobian is the prefactor of $dS$ when changing coordinates. Typically the Jacobian is memorised for popular coordinate systems, so you would just look up that $d\mathbf{S} = \mathbf{n}r^2\sin\theta d\phi d\theta$ on the surface of a sphere, in spherical coordinates. Here $\mathbf{n} = \mathbf{a}_r$ is the unit normal $(\sin \theta \cos \phi, \sin \theta \sin \phi,\cos \theta)$.

The integral is $-2.4\int \mathbf{a}_z d\mathbf{S}$ = $\int_{\phi = 0}^{2\pi} -2.4\int_{\theta = 0}^{\frac{\pi}{2}} (\mathbf{a}_z . \mathbf{a}_r) r^2\sin\theta d\theta d\phi$

I understand $a_z := (0,0,1)$ in Cartesian coords, and $\mathbf{a}_r := (\sin \theta \cos \phi, \sin \theta \sin \phi,\cos \theta)$ and so the dot product of the two vectors is $\mathbf{a}_z . \mathbf{a}_r = \cos\theta$.

Now, we have the integral over the upper hemisphere $(\theta,\phi) \in (0,\pi/2)\times(0,2\pi)$:

$\int_{\phi = 0}^{2\pi} \Big( \int_{\theta = 0}^{\frac{\pi}{2}} \cos \theta r^2\sin\theta d\theta \Big) d\phi$ = $r^2 \int_{\phi = 0}^{2\pi} \Big( \int_{\theta = 0}^{\frac{\pi}{2}} \cos \theta \sin\theta d\theta \Big) d\phi$ = $\frac{r^2}{2} \int_{\phi = 0}^{2\pi} \Big( \int_{\theta = 0}^{\frac{\pi}{2}} \sin 2\theta d\theta \Big) d\phi$

= $\frac{r^2}{2} \int_{\phi = 0}^{2\pi} \Big( \int_{\theta = 0}^{\frac{\pi}{2}} \sin 2\theta d\theta \Big) d\phi$ = $\frac{r^2}{4} \int_{\phi = 0}^{2\pi} \Big( \cos 0 - \cos \pi d\theta \Big) d\phi$ = $\frac{r^2}{2} \int_{\phi = 0}^{2\pi} 1 d\theta d\phi$ = $r^2 \pi$

All steps can be multiplied by $-2.4$ giving $-2.4\pi r^2$

(That's if I've made no errors)

• I know Jacobian for spherical coordinates but I don't understand how to find unit vector of surface. In this case $a_r$ is unit vector of surface. Also I don't understand why $a_z a_r=cos\theta$. Thank you for your answer. – Lamija37 Dec 17 '17 at 19:58
• By looking at a picture, the unit normal of a sphere at point $(x,y,z)$ is $\frac{1}{r}(x,y,z)$ (proportional to the position vector) The z component of the unit vector is $cos \theta$ in these coords. I still don't know what $a_z$ is. – Christian Fieldhouse Dec 17 '17 at 20:07
• My J vector is $\mathbf J=-2.4\mathbf k$ and in my question I wrote $J=-2.4\mathbf a_z$ instead of $\mathbf k$ – Lamija37 Dec 17 '17 at 20:09
• Ok got it. The z component of the unit normal is $cos \theta$ – Christian Fieldhouse Dec 17 '17 at 20:11