# Math Help - Vector Calculus in Spherical Polar Co-ords.

1. ## 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.

thanks.

2. 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 ...