Could someone help me with this problem? Evaluate $\int_C \mathbf{F} \cdot d\mathbf{r} $ where $$\mathbf{F} = \langle x^2, y^4-x , z^2 \sin z \rangle$$ and $C$ is the boundary if the portion of the surface: $z = x^2+y^2$, below $z=4$.
a) directly;
b) Using Stokes' Theorem.
How do I parameterize the boundary? And then what do I do with it? And how do I find $\mathbf {\hat n} $ and $ d\mathbf{S}?$
Edit: Can someone confirm or correct the answer $-4\pi$?
Upon direct parametrization we obtain $x = 2 \cos(t)$, $y =2 \sin(t)$ and $z=4$. This yields $${\mathbf r} (t) = (2\cos(t), 2 \sin(t), 4)$$ and $${\mathbf r}'(t) = (- 2 \sin(t), 2 \cos(t), 0).$$ Since $${\mathbf F}({\mathbf r}(t)) = (4\cos^2(t),16 \sin^4(t) - 2 \cos(t), 16 \sin(4)),$$ we see that $${\mathbf F}({\mathbf r}(t)) \cdot {\mathbf r}'(t) = - 8 \cos^2(t) \sin(t) + 32 \sin^4(t) \cos(t) - 4 \cos^2(t).$$ The integrals of the first two terms vanish, and a double-angle substitution yields $$\int_C {\mathbf F} \cdot d {\mathbf r} = - 4 \pi.$$ Using Stokes's theorem, notice that $$\nabla \times {\mathbf F} = (0,0,-1).$$ The natural function parametrization for the surface yields a normal $${\mathbf n} = \frac{(-2u,-2v,1)}{\sqrt{1+4u^2 + 4v^2}}.$$ The parametrization is $${\mathbf r}(u,v) = (u,v,\underbrace{u^2+v^2}_{f(u,v)}).$$ The normal vector is then $${\mathbf n} = \frac{(-f_u, -f_v, 1)}{\sqrt{1 + (f_u)^2 + (f_v)^2}}.$$ Therefore $$\iint_S \nabla \times {\mathbf F} \cdot d {\mathbf S} = \iint_A (-1) \, dA = - 4 \pi.$$ This can be seen by noticing that $d {\mathbf S} = \sqrt{1+4u^2+4v^2} \, dA$, which cancels out the denominator. Since the curl will take only the last coordinate and reverse the sign, the effect is that the surface integral is equal to minus the area of the projected region.