# Stokes' Theorem/Line Integral Question

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 $-4\pi$?

• The intersection of the surface $z = x^2 + y^2$ and the plane $z = 4$ is $x^2 + y^2 = 4$. Can you parameterise this? – Michael Albanese Jan 2 '15 at 4:18
• yes - $x= 2 \cos t \ y=2 \sin t$ z=4 - but do I just plug those into the integral? If I do I get $\frac{8}{3}-4\pi$ – Gabriel Jan 2 '15 at 4:26
• I'm having trouble with the Stokes' theorem part (also because if the orientation of C is counterclockwise, the corresponding orientation on the surface is downward and I'm not sure what to do with that) – Gabriel Jan 3 '15 at 23:27
• The answer is indeed $-4 \pi$. If you arrived at $\displaystyle \frac{8}{3} - 4 \pi$ for your line integral you have probably miscalculated $$\int_0^{2 \pi} - 8 \cos^2(t) \sin(t) \, dt.$$ – Mark Fantini Jan 4 '15 at 0:00

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.
• @Gabriel Yes, all normals in vector theorems are unit. Usually, though, the length gets canceled by multiplication with $dS$, which is the length of the normal vector times the area element of the projected region. If this answer helped you, consider accepting it. – Mark Fantini Jan 4 '15 at 2:02
• And the normal was ${\mathbf n} = \frac{(-f_u, -f_v, 1)}{\sqrt{1 + (f_u)^2 + (f_v)^2}}$ instead of ${\mathbf n} = \frac{(f_u, f_v, -1)}{\sqrt{1 + (f_u)^2 + (f_v)^2}}$ because the orientation was downward so you need the negative normal? – Gabriel Jan 4 '15 at 15:49