Use Stokes’ Theorem to evaluate $\displaystyle \int_c^.F \cdot dr $

C: is the boundary of the portion of $\displaystyle z = x^2 + y^2 $ below z = 4, oriented downward,

$\displaystyle F = <x^2, y^4 - x, z^2sin z> $

Can someone please check over my work thanks!

First I found N --> $\displaystyle N = \frac{<2x, 2y, -1>}{\sqrt{4x^2 + 4y^2 + 1^2}} $

Then i found $\displaystyle \bigtriangledown \times F = <0, 0, -1> $

from using the formula $\displaystyle \bigtriangledown \times F = <F_{3y} - F_{2z}, -F_{3x} + F_{1z}, F_{2x} - F_{1y}> $

Now $\displaystyle \bigtriangledown \times F \cdot N = \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}} $

Then $\displaystyle dS = {\sqrt{4x^2 + 4y^2 + 1}}dA $

So putting it all together....

$\displaystyle \int\int_R^. \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}} \cdot \sqrt{4x^2 + 4y^2 + 1} $

The square roots cancel and reduce to....

$\displaystyle \int\int_R^. 1 dA $

This double integral of 1 dA is equal to the Area of the Region

And using $\displaystyle z = x^2 + y^2 $ and $\displaystyle z = 4 $ we come up with $\displaystyle r^2 = 4 .... r = 2 $

$\displaystyle A(R) = \pi r^2 = \pi(2^2) = 4\pi $

Does $\displaystyle 4\pi $ look correct for the answer?

Thanks for looking