Undergrad Gauss's law of sphere using integral

[pinkfishegg]

Hey I was just practicing Gauss's law outside a sphere of radius R with total charge q enclosed. So I know they easiest way to do this is:

∫E⋅da=Q/ε
E*4π*r^2=q/ε
E=q/(4*πε) in the r-hat direction

But I am confusing about setting up the integral to get the same result

I tried
∫ 0 to pi ∫0 to 2pi E*r^2sin(Φ) dΦdΘ for the LHS

i got this from the front of Griffiths where it says:
dl=dr(r-hat)+sdφ d(θ-hat)+rsin(θ)dφ (φ-hat)

But i keep getting 0 when i take this integral sin i get -cos(2π)-cos(0)= -1-(-1)=0

So there's obviously something wrong with my setup. I need to integrate over θ and φ because they are on the surface but I'm not sure about the terms in front. Aren't they supposed to be they ones in front from the dl term?..

Trying to practice this to be able to answer harder questions :P

 

[Charles Link]

The surface integral is not a $d \vec{l}$ type integration. The surface differential is $dA=r^2 sin(\theta) \, \, d \theta \, d \phi$ in polar coordinates. $\theta$ gets integrated from $0$ to $\pi$, and $\phi$ from $0$ to $2 \pi$. The integrals are separable, and there is complete symmetry as $E$ is assumed to be a constant, (depending only on $r$), independent of $\theta$ and $\phi$. The result of these two integrals is $4 \pi$.

 

[pinkfishegg]

ohhh so i got the bounds for φ and θ mixed up..

 

[pinkfishegg]

ohhh so i got the bounds for φ and θ mixed up..

So Φ is from top to bottom and θ from left to right correct? why are those the bounds. It seems like it could go either way and you'd still get the entire sphere.

 

[Charles Link]

So Φ is from top to bottom and θ from left to right correct? why are those the bounds. It seems like it could go either way and you'd still get the entire sphere.

$\phi$ is the azimuthal angle, basically counterclockwise, and goes all the way around (360 degrees). $\theta$ is the polar angle, and when it points all the way downward the angle is 180 degrees. When the point lies in the x-y plane, the polar angle is 90 degrees. $\\$ Additional item, is that spherical coordinates is not a system that you can draw about some origin, (at radius $r$ away from some other origin), where for small angles $\phi$ is left and right (x-direction), and $\theta$ is the elevation angle (y-direction) . This type of coordinate system is also used at times for sources that have a narrow beam spread, but it is not spherical coordinates.

 

[pinkfishegg]

$\phi$ is the azimuthal angle, basically counterclockwise, and goes all the way around (360 degrees). $\theta$ is the polar angle, and when it points all the way downward the angle is 180 degrees. When the point lies in the x-y plane, the polar angle is 90 degrees.

ok i just looked at this: https://en.wikipedia.org/wiki/Spherical_coordinate_system. sometimes its opposite in math so that's why i mixed it up.

 

[Charles Link]

Please see also my edited comment (Additional item) in my previous post.