Thread: Vector Calculus in Spherical Polar Co-ords.

Sorry the title is meant to be in cylindrical co-ords, my bad.
Hello, first post from me.

My problem is to do with the divergance thereom, i have a vector field F.

F(x,y,z) = ( (x^2+y^2)^3 (-yi + xj +zk) ) / ( [1 + (x^2 + y^2)^4 ]^4)

Sorry for the way it is layed out, i do not know how to use the math symbols.

and surface theta = x^2 + y^2 =< 1 for 0 =< z =< 3 (cynlinder from xy plane to z=3 plane)

Then i have to calculate using the divergance theroem,
integral: (nubla.F)dV over volume of theta.

Which is equal to
integral: F.n ds over surface of theta.
(where n is the unit vector perpendicular to the surface)
I need to calculate the bottom identity:

So i split the surface into 3 seperate surfaces (pretty sure i can do this)
where i calculate integral of top circle + integral of bottom circle + integral of strip that goes around the cylinder.

Switching to polar co-ords the first 2 circles, top and bottom cancel out (as unit normal vector is k, in the first one and -k in the second, due to the dot product have property a.-b = -a.b so +1 and -1 can be treateed as constants which go to the outside of the intergrand and end up with +int(F.k ds) -int(F.k dS) which is zero, as ds = ds)

So im left with the strip, switching the F field to polar co-ords yields
F= 1/16 e(angluar) + z/16 e(z) where z is unit vector for height, (ie perpendicular to both y and x axis).

No this has no component for e(p) where p is the unit vector point at right angles to the unit vector for angle and unit vector for height.

so the dot product would yeild 0.

Im pretty sure this is incorrect, please help
thanks.

Hello !

Please consider my post with care since I am not a specialist !
What I can say is that I agree with you for what concerns the strip (integral = 0) but not for the 2 circles.

For the bottom circle, you must integrate F.n ds on the surface defined by x²+y² < 1 and z=0. You have F.n=0 because n=-k and F has coordinate 0 on vector k (z=0)

For the top circle, you must integrate F.n ds on the surface defined by x²+y² < 1 and z=3. You have F.n ds = (3(x^2+y^2)^3 / ( [1 + (x^2 + y^2)^4 ]^4)

I hope that I am not wrong ...