Use Gauss's Law to find the charge contained in the solid hemisphere

[RedDelicious]

Homework Statement

edit: I had put this in the calculus section because it was a problem from Stewart but I guess it's closer to a physics problem considering the use of Gauss's Law. My apologies for any confusion this my
[/B]
I've been trying to do this problem without making use of the divergence theorem but have been doing something wrong and cannot figure it out.

Use Gauss’s Law to find the charge contained in the solid hemisphere

x^2+y^2+z^2\le a^2,\:z\ge 0

if the electric field is

E\:=\:<x,\:y,\:2z>

The Attempt at a Solution

[/B]
I use the parametric equation for a sphere to model the hemisphere, with the idea that only the limits of integration over the parameter domain at the end will change, but the parameterization is the same.

r\left(\phi ,\theta \right)=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:acos\phi >

and using this to parameterize the electric field for the hemisphere

F=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:2acos\phi >

Using

\int \int F\cdot dS\:=\:\int \int F\cdot \left(r_u\:x\:r_v\right)\:dA

I get

F\cdot \left(r_u\:x\:r_v\right)\:=a^3sin^3\phi cos^2\theta +a^3sin^3\phi sin^2\theta +2a^3sin\phi cos^2\phi =a^3sin\phi cos^2\phi

\int _0^{2\pi }\int _0^{\frac{\pi }{2}}a^3sin\phi cos^2\phi \:d\phi d\theta

=\frac{2}{3}\pi a^3

which is a factor of 4 off, as the answer is =\frac{8}{3}\pi a^3\gamma

*not including the electric constant

Thanks,
RedDelicious

 

[blue_leaf77]

I get
$$
F\cdot \left(r_u\:x\:r_v\right)\:=a^3sin^3\phi cos^2\theta +a^3sin^3\phi sin^2\theta +2a^3sin\phi cos^2\phi =a^3sin\phi cos^2\phi
$$

This is incorrect. For instance, how will you simplify the first and second term in the middle expression? Moreover, I wonder how come the power of $a$ is 3 not 2.

 

[RedDelicious]

This is incorrect. For instance, how will you simplify the first and second term in the middle expression? Moreover, I wonder how come the power of $a$ is 3 not 2.

Thank you, I wasn't sure if it'd be needed at first, but since I did something wrong here is what I did step by step,

The parametric equation for the sphere is

r\left(\phi ,\theta \right)=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:acos\phi >

Using this to parameterize the electric field,

E=\:<asin\phi cos\theta ,\:asin\phi sin\theta ,\:2acos\phi >
By an example in the book, for a sphere,

r_{\phi }\:x\:r_{\theta }\:=\:<a^2sin^2\left(\phi \right)cos\left(\theta \right),\:a^2sin^2\left(\phi \right)sin\left(\theta \right),\:a^2sin\left(\phi \right)cos\left(\phi \right)>

and so

E\cdot \left(r_{\phi }\:x\:r_{\theta }\right)\:=a^3sin^3\phi cos^2\theta +a^3sin^3\phi sin^2\theta +2a^3sin\phi cos^2\phi<br /> \\<br /> =a^3sin^3\left(\phi \right)\left[cos^2\left(\theta \right)+sin^2\left(\theta \right)\right]+2a^3sin\left(\phi \right)cos^2\left(\phi \right)<br /> \\<br /> =a^3sin^3\left(\phi \right)+2a^3sin\left(\phi \right)cos^2\left(\phi \right)<br /> \\<br /> =a^3sin\left(\phi \right)\left(sin^2\left(\phi \right)+2cos^2\left(\phi \right)\right)\\<br /> =a^3sin\left(\phi \right)\left(sin^2\left(\phi \right)+cos^2\left(\phi \right)+cos^2\left(\phi \right)\right)\\<br /> =a^3sin\left(\phi \right)\left(1+cos^2\left(\phi \right)\right)

which when substituted into the integral does give me the correct answer of 8a^3pi/3 I repeatedly made a stupid error in simplifying that (1+cos^2(phi)) as just cos^2(phi). Unless you believe that I've erroneously arrived at the answer, I think all is well. Thank you for bringing that step to my attention