# Surface integral for line current

1. Nov 15, 2015

### Incand

1. The problem statement, all variables and given/known data
Calculate the integral $\oint_C \vec F \cdot d\vec S$, where $C$ is the closed curve constructed by the intersection of the surfaces $z = \frac{x^2+y^2}{4a}$ and $x^2+y^2+z^2=9a^2$, and $\vec F$ is the field $\vec F = F_0\left( \frac{a}{\rho}+\frac{\rho^2}{a^2} \right)\hat \phi$. ($a$ and $F_0$ are constants.)

(I'm guessing this fits better here than in the calculus forum since since the question is as preparation for EM-class)
2. Relevant equations
Line current with strength $J$
$\vec F = \frac{J}{2\pi \rho} \hat \phi$.

3. The attempt at a solution
We see the first part of the force is a line current with strength $2\pi F_0a$ so it's total contribution to the integral over a closed loop is $2\pi F_0a$.

I'm left with $\vec F_2 = F_0\frac{\rho^2}{a^2}\hat \phi$. We note from $z=\frac{x^2+y^2}{4a}$ that if $a >0$ $z$ has to be strictly positive and if $a < 0$ $z$ is strictly negative.
Substituting into the second equation we have
$(z+2a)^2-4a^2 = 9a^2 \Longrightarrow z=(\pm \sqrt{13} -2)a$. We conclude that $z$ is constant and since $\rho^2 = x^2+y^2$, $\rho$ is also constant.
Calculating the curl we get $\nabla \times \vec F_2 = \frac{3F_0}{a^2}\rho \hat z$.
Using stokes theorem and since we know $\rho$ is constant and using $\hat z$ as the normal vector
$\int_C \vec F_2 \cdot d\vec r = \frac{3F_0}{a^2}\rho \int_S dS = \frac{3\pi F_0}{a^2}\rho^3$
Let's assume that $a>0$ we then have that $\rho = 2a\sqrt{\sqrt{13}-2}$. Using this we have
$\int_C \vec F_2\cdot \vec r = 24\pi F_0a\left( \sqrt{13}-2 \right)^{3/2}$.
So the total value of integral should be
$2\pi F_0a\left(12(\sqrt{13}-2)^{3/2}+1\right)$.

Another approach would be to use the line element $d\vec r = d\rho \hat \rho + \rho d\phi \hat \phi + dz \hat z$
$\int_C \vec F_2 \cdot d\vec r = \int_C \frac{F_0\rho^3}{a^2}d\phi$ And everything inside should be a constant. Since both $z$ and $\rho$ are constant we should be on a circle thats parallel to the $xy$-plane and we should integrate from $\phi =0 \to \phi = 2\pi$. So the integral becomes
$\int_C \vec F_2 \cdot d\vec r = \frac{2\pi F_0 \rho^3}{a^2} = 8\pi aF_0(\sqrt{13}-2)^{3/2}$. Giving us the value
$2\pi F_0a\left(8(\sqrt{13}-2)^{3/2}+1\right)$.

According to the answer the second approach gave me the right answer and the first one is wrong but I can't see where I go wrong?

2. Nov 15, 2015

### RUber

Have you drawn a picture of what the contour looks like?
This is the intersection of a cone and a sphere.
Since your integral uses spherical coordinates, convert the contour to spherical coordinates as well. That should help.

3. Nov 15, 2015

### Incand

Sorry I should've clarified better. The field is given in cylindrical coordinates (the book uses $\rho$ for cylindrical.)
Should be a paraboloid not a cone since we have $z=\frac{x^2+y^2}{4a}$ not $z^2=\frac{x^2+y^2}{4a}$ so the intersection should be a circle. I did make a drawing but I'm not too great at those. Do you think the second solution is correct or did I just happen upon the correct answer there? Be cause I feel I did most of the work similar for both approaches.

4. Nov 15, 2015

### RUber

Yes, you are z is growing with the square of the radius of the circle, not linearly with the radius like in a cone.
Your contour is a circle parallel to the xy plane. So the second method leaves fewer opportunities to make a mistake.

5. Nov 20, 2015

### Incand

I realised what my error was here. When I calculated the surface integral i got from stokes theorem I had
$3F_0a^2\int_S \rho d\vec S$. S is a $z$-surface so the surface element is given by $d\vec S = \hat z \rho d\rho r\phi$. Using this we have (let $\rho_0$ be the radius of the circle as to not confuse them)
$3F_0a^2 \int_0^{\rho_0}\int_0^2\pi \rho^2 d\phi d\rho = 2\pi F_0\rho_0^3$ yielding the correct answer in the first case as well. So the mistake I made was that I didn't accountfor that I have scale factors in curvilinear coordinates.