Any help would be MUCH appreciated! 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!