Any help would be MUCH appreciated!(adsbygoogle = window.adsbygoogle || []).push({});

Vector A=yi+zj+xk

Surface S is defined by z=1-x^2-y^2 where z is greater or equal to 0.

Find the value of the surface integral of (grad x A)*dA

(The “x” is the cross product.)

I know this uses Stoke’s Theorem: line integral of the closed curve of A*dS = surface integral (grad x A)*n*dA. So, the line integral of vector A around the closed curve is equal to the curl of A over the surface defined by the curve.

I’m fairly sure I’ve set everything up correctly. I have to find (gradxA). Then I find the unit normal vector by taking the gradient of the surface and dividing it by its magnitude. Then I find dS by making it (dx*dy)/(the magnitude of n*k). Then I do the dot product of my (gradxA) vector and my unit normal vector. Multiply by the magnitude of n*k, then the double integral is set up. I think I have to switch to polar coordinates, at the end.

I figured out (grad x A) by using the determinant. It's -i+j-k.

Now I need to find the normal of the given surface. For this, I take the gradient of the surface. I rearranged the surface so that

x^2+y^2+z=1. The gradient is 2xi+2yj+k. I need the unit vector of the normal, so I divide by the magnitude. This is where I get messed up, and unsure if I'm doing this right. I get the square root of (4x^2+4y^2+1). There's no way this is right, because almost nothing will simplify...

I hope I'm not completely wrong. Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Stoke's Theorem

**Physics Forums | Science Articles, Homework Help, Discussion**