I am trying to solve the following question, which is a standard Surface Integral/Stokes' theorem question. Unfortunately, I've tried to calculate it several times, and each time get the answer slightly wrong.
I calculated $\nabla \times B = 3(0,0,x^2+y^2)$
The normal vector for the 'lid' of the shape will be $(0,0,1)$, and for the bottom $(0,0,-1)$
As $\nabla \times B$ only has a non-zero z component, we only need to calculate the z-component of the normal vector on the 'side' of the shape. I calculated this to be $r$, where I used cylindrical co-ordinates to describe the shape in terms of $r$ and $\theta$, as $(r\cos(\theta), r\sin(\theta), r^2)$.
Then using stoke's theorem, I calculated the line integrals to come out as $-\frac{6\pi}{4}(1+\frac{1}{3^4})$, but calculating the surface integral directly I got the top component to come out as $2\pi$, the bottom component to come out as $-6\pi\frac{1}{3^4}$ and the middle component to come out as $\frac{6\pi}{4}(1-\frac{1}{3^4})$
That means I found the surface integral to come out as $\frac{4}{9}\pi$ and the line integral to come out as $\frac{41}{27}\pi$ which is not the same :(
In any case, I have spent a very long time re-doing the calculations and have made no progress, so thank you very much