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# Surface integral Watch

1. I need to evaluate the integral

where A = (x, y, z) and S is the part of the cylinder , with the limits , , .

My working:

First, we need to find an expression for the unit normal for the surface .

I did this by using the expression where f = f(x,y,z) represents some function with a constant surface, i.e. .

So and . Thus .

We also know that for a cylinder, the surface element dS is given by , and in cylindrical polars, and . This gives:

and .

Taking the dot product of the two vectors above, we get:

.

Our final expression is then

If I put in rho = 4 (from the original equation of the cylinder, I'm assuming it's of the form x^2 + y^2 = rho^2?) and limits of 0 to 5 for z and 0 to 2pi for phi... I don't get the required answer of 40pi. Where have I gone wrong??? If I had a rho instead of a rho^2 inside of my integral I'd get the right answer...

2. I think your simplification of the normal vector may be wrong.

EDIT: Sorry, I see you cancelled the radium out there. My mistake.
3. it says the x,y,z limits are greater than zero so maybe your new limit should be pi rather than 2pi.
4. i did it from start to finish and also get 160pi.

it definitely is 160pi. the flux density through the surface is 4 at any point on the outside of the cylinder (z component doesnt matter as its not normal to the surface) as the field is 4 and pointing normal to the surface. the area of the surface is 5 X 2 x pi x 4 is 40pi. so its defo 160pi

pdzd(theta) is a description of the surface area and when integrated with your limits gives the correct surface area.

the second p is your flux normal to this surface at this surface, and p is constant at your surface.

mm aahh yes limits hurrr, that makes sense
5. (Original post by djpailo)
it says the x,y,z limits are greater than zero so maybe your new limit should be pi rather than 2pi.
But then I'd get 80pi, which is still twice the required answer Any other tips? I get the feeling that perhaps I've evaluated the normal vector or the dot product improperly.
6. Oooh actually... the limits should be zero to pi/2! It says x > 0 and y >0 ie cos(phi) > 0 and sin(phi) > 0 - this is only valid in the range 0 < phi < pi/2. We're only considering the cylinder in the first positive quadrant basically. This will give me the required answer of 40pi. Thanks guys (and to superfoggy too)
7. (Original post by trm90)
Oooh actually... the limits should be zero to pi/2! It says x > 0 and y >0 ie cos(phi) > 0 and sin(phi) > 0 - this is only valid in the range 0 < phi < pi/2. We're only considering the cylinder in the first positive quadrant basically. This will give me the required answer of 40pi. Thanks guys (and to superfoggy too)
haha awesome

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