# Integration using Stokes' theorem

I have a problem that I can't seem to figure out. Given a surface $$S$$: $$\begin{cases} x^2+y^2 \leq 1 \newline z = y^2 \end{cases}$$ Let $$C$$ be the edge of $$S$$. $$C$$ is oriented so that the projection of $$C$$ on the $$xy$$-plane runs counter-clockwise. Calculate: $$\begin{equation} \oint_{C}y^2dx + xy^2 dy+ xzdz \end{equation}$$

1. Directly
2. Using Stokes's theorem

I've tried to make a sketch of the area with the orientation:  I've at least gotten $$\begin{equation} \vec{F} = y^2 \vec{i} + xy^2 \vec{j} + xz \vec{k} \end{equation}$$ and $$\begin{equation} curlF = -z\vec{j} + (y^2-2y)\vec{k} \end{equation}$$

For calculating the integral directly I've tried a parametrization: $$\begin{cases} x = r\cos(t) \newline y = r\sin(t) \newline z = y^2 = r^2\sin^2(t) \end{cases}$$ with $$0 \leq r \leq 1$$ and $$0 \leq t \leq 2\pi$$. This gave me: $$\begin{equation} \int_{0}^{1} \int_{0}^{2\pi} r^2\sin^2(t) + r^2\cos(t)\sin^2(t) + r^3\cos(t)\sin^2(t) dtdr = \frac{1}{3} \pi \end{equation}$$

But now for $$\iint_{R} curlF \cdot N dS$$ I can't figure out how to proceed. I can't seem to figure out what $$N$$ is and what integration bounds I need to use. The examples in my book and my lecture aren't all that similar and I don't see how to apply those to this situation. If someone could give me a hint on how to proceed I'd be very grateful.

• Your parametrization of $C$ does not make sense, it should be is a line.. If you let $r$ tkae any value in [0,1] your are parametrising the surface, not the border. $$x = \cos t,\quad y = \sin t, \quad z = \sin^2 t$$ Using the parametrization the line integral becomes a 1d integral. Jun 2 at 16:53

As regards the other side of Stokes' theorem, we may consider the surface $$S=\{(x,y,y^2): x^2+y^2\leq 1\}$$ oriented upwards. A parametrization of $$S$$ is given by $$\mathbf{r}(x,y)=(x,y,y^2) \; \text{with (x,y) such that x^2+y^2\leq 1} \implies \mathbf{r}_x\times \mathbf{r}_y= (0,-2y,1).$$ Therefore $$\iint_S \text{curl}(\mathbf{F})\cdot d\mathbf{S}=\iint_{\{x^2+y^2\leq 1\}} (0,-z,y^2-2y)\cdot (0,-2y,1) \,dx dy=\frac{\pi}{4}.$$ which is equal (please fill the details) to the direct computation made by PierreCarre.
Regarding the direct calculation, a possible parametrization would be $$\begin{cases} x = \cos t\\y = \sin t\\ z = \sin^2t \end{cases}$$
The line integral can be computed as $$\int_C y^2 dx + xy^2dy+xz dz = \int_0^{2 \pi} (\sin^2 t \cdot (-\sin t)+\cos t \sin^2 t \cdot \cos t + \cos t \sin^2 t \cdot 2 \cos t \sin t) dt = \frac{\pi}{4}$$