[Calc 3] Gauss' Law for Elecricity over a hemisphere

[mjdiaz89]

Hello,

Thank you very much for takign the time to help me out. This is an applied calculus 3 (multivariable calc) level problem. I have an answer, but it is nonsensical as the units are not in accordance with units of charge, C.

Homework Statement

Instructions: Use Gauss' Law for electricity and the relationship q=\int\int\int_{Q} \rho dV.

Question: For E=\left\langle 2xz^{2},2yx^2,2zy^2\right\rangle, find the total charge in the hemisphere z=\sqrt{R^{2}-x^{2}-y^{2}.

Homework Equations

\nabla \bullet E=\rho/\epsilon_{o}

The Attempt at a Solution

Finding the divergence of E:
\nabla \bullet E = \frac{\partial 2xz^{2}}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}
Subsituting z=\sqrt{R^{2}-x^{2}-y^{2} in the partial of x as z is a function of x:
= \frac{\partial 2x(R^{2}-x^{2}-y^{2})}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}
Yeilds:
\nabla \bullet E = 2(R^{2}-2x^{2})

Now, this means that \rho = 2 \epsilon_o (R^{2}-2x^{2})

Using the relation q=\int\int\int_{Q} \rho dV:
and setting up the region of integration:
q = 2 \epsilon_o \int^{R}_{-R} \int^{\sqrt{R^{2}-x^{2}}}_{-\sqrt{R^{2}-x^{2}}} \int^{\sqrt{R^{2}-x^{2}-y^{2}}}_{0}(R^{2}-2^{2}) dz dy dx

Seeing spherical symmetry, a conversion from rectangular to spherical coordinates will be convenient. *NOTE: My book and teacher use phi to indicate the angle from the z-axis and theta as the angle from the x-axis; like this: http://mathworld.wolfram.com/SphericalCoordinates.html * :
R^{2} - x^{2} = \rho^{2} - (\rho sin\phi cos\theta)^{2}

Switching limits of integration and substituting \rho dV

q = 2 \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} \int^{R}_{0} \rho^{2} sin\phi (\rho^{2} - (\rho sin\phi cos\theta)^{2}) d\rho d\phi d\theta
Distributing and integrating d\rhp
q = 2 \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} [\rho^{5}/5sin\phi - 2\rho^{5}/5sin^{3}\phi cos^{2}\theta]|^{R}_{0} d\phi d\theta

After integrating wrt d\rho and moving R^5 outside of the iterated integrals, the iterated integral remains as (Notice the weird R^5 units!):
q = 2/5 R^{5} \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} [sin\phi - 2sin^{3}\phi cos^{2}\theta] d\phi d\theta

Once all the trigonometric integrations are done (which is just as long as the work up to this point), my finals answer is
\frac{4}{15} \pi R^{5} \epsilon_o

The units of this answer are [R^{5}][\epsilon_o] = m^{5} * \frac{C^{2}}{Nm^{2}}\neq C

Im sure there is an error I made as this is a law. However, my teacher was unable to find my error after visiting him in his office. I have been thinking about this problem for a few days now and ...well... I am still without a sensical answer. Any help?

 

[gabbagabbahey]

Question: For E=\left\langle 2xz^{2},2yx^2,2zy^2\right\rangle, find the total charge in the hemisphere z=\sqrt{R^{2}-x^{2}-y^{2}.

You should take note here that for E to represent an actual electric field, it must have units of N/C. The only way this can happen is if you assume (sort of hand-wavingly) that there is some constant k=1 \frac{\text{N}}{\text{m}^3\text{C}} multiplying this expression. If you were to assume this, then when you found your answer to be proportional to \epsilon_0 R^5, you would have that it is actually proportional to k\epsilon_0 R^5 which would give you the correct units of charge. However, since this is a math course and not a physics course I doubt they expect you to correct the units of E in this manner.

There is a mistake on your part though:

\nabla \bullet E = \frac{\partial 2xz^{2}}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}
Subsituting z=\sqrt{R^{2}-x^{2}-y^{2} in the partial of x as z is a function of x:

When you do this substitution, you are effectively assuming that z=\sqrt{R^{2}-x^{2}-y^{2} everywhere in the hemisphere; but that is only true on the surface of the hemisphere. At a general point in the hemisphere, z is independent of x and y, so \frac{\partial }{\partial x} (2xz^2)=2z^2 and the divergence of E becomes 2z^2+2x^2+2y^2=2\rho^2.