Stokes' theorem over 3D object

Question

Could anyone please help me with the interpretation of the following question?

Question:

S is the surface $z=x^2+y^2$ bounded by the planes $z=0$ and $z=4$. Verify Stokes' theorem for a vector field $\boldsymbol F=xy \boldsymbol i+x^3 \boldsymbol j + xz \boldsymbol k$

So my task is to show that:

$\int_scurl \, \boldsymbol F.dS = \oint_c \boldsymbol F. d \boldsymbol r$

starting with the RHS I evaluate $\oint_c \boldsymbol F. d \boldsymbol r$ around the boundary of the circle $x^2 + y^2 = 4$ in the plane $z=4$ and I get the answer $12 \pi$ which agrees with the text book I am using.

My question is, for the LHS, is the surface over which I am expected to integrate the one defined by the object that looks like a box with a paraboloid in the top?

So I'll have to integrate over the five planar faces, the paraboloid and the four corners of the z = 4 plane left by the removal of the circle $x^2 + y^2 = 4$?

Edit: Basically, re-reading the question, I see that I invented the "box" object. To apply Stokes' theorem I need an open surface bounded by a simple curve, but what is the open surface in this question? I see that the text book expects that the simple curve bounding it is the circle $x^2+y^2=4$ in the z = 4 plane (I inferred this from my answer being correct) but I can't see how the surface is defined.

Further, I'm not fully sure how to integrate over the paraboloid. Can anyone provide pointers there, or a link to a similar example?

Thanks, Mitch.

Answer 1

As you found out yourself you have to integrate solely over the paraboloid $P$. Use the parametrization $$P:\quad (r,\phi)\mapsto \bigl(r\cos\phi,\>r\sin\phi, \>r^2\big)\qquad(0\leq r\leq 2, \ 0\leq\phi\leq 2\pi)\ .$$ Make sure that the induced orientation of $P$ corresponds with the chosen orientation of the boundary circle.

Answer 2

Ok, I understand now, except for one thing:

$\int_scurl \boldsymbol F .d\boldsymbol S= \int_scurl \boldsymbol F\,.\,\hat{\boldsymbol n}\, dS$

where $curl \boldsymbol F = -z \boldsymbol j + (3x^2-x) \boldsymbol k$

and $\hat{\boldsymbol n} = \frac{\nabla S}{\left| \nabla S\right|} = \frac{2x \boldsymbol i + 2y \boldsymbol j - \boldsymbol k}{\sqrt{4x^2 + 4y^2+1}}$

So I need to calculate:

$\int_s (-z \boldsymbol j + (3x^2-x) \boldsymbol k) . \frac{2x \boldsymbol i + 2y \boldsymbol j - \boldsymbol k}{\sqrt{4x^2 + 4y^2+1}} dS$

Is that correct?

If I leave out the term $\sqrt{4x^2 + 4y^2+1}$ I get the right answer, but should it not be included?

Thanks Mitch.