# Finding a closed line integral using Stokes' Theorem

Find the line integral $\int_C \vec{F} \cdot \vec{dS}$, where $C$ is the circle of radius 3 in the $xz$-plane oriented counter-clockwise when looking from the points $(0, 1, 0)$ into the plane and where $\vec{F}$ is the vector field $$\vec{F}(x, y, z) = \langle 2x^2z + x^5, \cos(e^y), -2xz^2 + \sin(\sin(z)) \rangle$$

I got $\vec{r}(u, v) = \langle u \cos (v), 0, u\cdot \sin(v)\rangle$.

Then I calculated that $\text{curl}(\vec{F}) = \langle 0, 2x^2 - 2z^2, 0\rangle$ and $\text{curl}(\vec{F})(\vec{r}(u, v)) = \langle 0, -u, 0\rangle$.

Using Stokes' Theorem and taking the dot product I get the integral: $$\int_0^3\!\int_0^{2\pi} -2u^3(\cos^2v - \sin^2v) \,dvdu$$

Is that correct?

## 1 Answer

Hint: Numerically, the line integral is approximately

$$\oint_{C}\vec{F}\cdot\mathrm{d}\vec{r}\approx254.469.$$

But the integral you arrived at vanishes to zero identically:

$$\int_0^3\!\int_0^{2\pi} -2u^3(\cos^2v - \sin^2v) \,dvdu=0.$$

So clearly there's a problem with your integral. Specifically, you miscalculated the curl. The correct curl is:

$$\nabla\times\vec{F}=\langle 0,2(x^2+z^2),0 \rangle.$$

Can you correct the rest of your work there?