Current through an ampere loop I am trying to apply Ampere's law to find a magnetic field in coaxial cylinders. The inner cylinder with radius $2a$ and outer cylinder is $4a$. Current flows along positive $z$ for inner cylinder and negative $z$ for outer cylinder.
The magnetic field I am trying to find is at radius $3a$.
One thing i am confused on is how to calculate the total current through the surface for the integral at $3a$.
The answer given is 7/12 * I ${\mu_0}$  but I am not sure how they got 7/12.
I calculated the total current for the two cylinders as:  ${\mu_0  I (16\pi a^2-4\pi a^2)} = \mu_0I12\pi a^2$
And subtracted that by the total current up to radius 3a:  ${\mu_0I(9\pi a^2 -4\pi a^2) = \mu_0 I 5 \pi a^2 }$
This gives:
Current in surface up to radius 3a: ${\mu_0 I 7 \pi a^2 }$
Vastly different to the answer of ${7/12 * I \mu_0}$
What am i getting wrong here?
 A: You first need to calculate the current density, $J = \dfrac{\partial I}{\partial S}$.  For $r<2a$, you have $J_{\rm in} = \dfrac I{4\pi (2a)^2}$ whereas for $r>2a$ you have $J_{\rm out}=\dfrac{I}{4\pi (4a)^2}$.
Recall that $$I = \iint_S \mathbf{\vec J} \cdot \mathrm d\mathbf{\vec S}$$ so for $r<3a$, you've got $$I = 4 \pi (2a)^2 J_{\rm in}-4\pi \left[ (3a)^2 - (2a)^2\right]J_{\rm out} $$ which gives you your answer.
A: The current density ($J$) will be given by $J_{\text{in}}=\frac{I}{\pi(2a)^2}$
Similarily, $J_{\text{out}}=\frac{I}{\pi((4a)^2)-(2a)^2)}=\frac{I}{12\pi a^2}$ 
Hence at the distance of $r=3a$, Using Ampere's rule
$B\cdot 2\pi(3a)=\mu_{0}I_{\text{net}}=\mu_{0}(I_{\text{in}}-I_{\text{out}})$
$B\cdot 2\pi(3a)=\mu_{0}(\pi(2a)^2J_{\text{in}}-\pi((3a)^2-(2a)^2)J_{\text{out}})$ 
Hence,
$\text{Line integral of }B=\frac{7}{12}\mu_{0}I$
Which gives you the right answer
A: $$\int \vec{B} \cdot \vec{dl} = \mu_{0} I_{enc}$$
Letting positive z point out of the screen. At radius 3a, the total enclosed current from the inner cylinder, would be "I"
So,
$$\int \vec{B} \cdot \vec{dl} = \mu_{0} I + \mu_0 \iint \vec{J}_{2} \cdot \vec{da}_{2}$$
Because we have oriented our z axis out of the page, I know that the $\vec{da}_{2}$ element points opposite $\vec{J}_{2}$, So by definition of the dot product.
$$\int \vec{B} \cdot \vec{dl} = \mu_{0} I + \mu_0 \iint -|\vec{J}_{2}| |\vec{da}_{2}|$$
$$\int \vec{B} \cdot \vec{dl} = \mu_{0} I - \mu_0  |\vec{J}_{2}| \iint |\vec{da}_{2}|$$
We know that $|\vec{J}_{2}| = \frac{I}{Area}$
The outer cylinder, only carries a current I,  for a small strip, between a cylindrical radius $2a < \rho < 4a$
This area, is the area between circles of radius 4a, and radius 2a, which is $\pi(4a)^2 - \pi(2a)^2$
So the current density
$|\vec{J}_{2}| = \frac{I}{\pi(4a)^2-\pi(2a)^2}$
Substituting into our equation:
$$\int \vec{B} \cdot \vec{dl} = \mu_{0} I - \mu_0  \frac{I}{\pi(4a)^2 - \pi(2a)^2} \iint |\vec{da}_{2}|$$
Finally,what is the area in question for the second integral? Well We have summed up the current from the inner cylinder(0 to 2a), so what's left is the area between 2a and 3a,
Which is $\pi(3a)^2- \pi(2a)^2$
So
$\int \vec{B} \cdot \vec{dl} = \mu_{0} I - \mu_0  \frac{I (\pi(3a)^2- \pi(2a)^2)}{\pi(4a)^2 - \pi(2a)^2} $
$$\int \vec{B} \cdot \vec{dl} = \frac{7}{12} \mu_0 I  $$
