Vector Calc 12-25-2020

I solved this integral using Stoke's theorem which says:

$\int \int_{S} \vec{F} \cdot d\vec{S} =\int \int_{S} \left(\nabla \times \vec{G}\right) \cdot d\vec{S} = \oint_{C} \vec{G} \cdot d\vec{r}$

Here, the boundary of the open surface is a circle centered at the origin having a radius of $2\pi$ . In other words, the boundary can be paramererised as such:

$\vec{r} = x \ \hat{i} + y \ \hat{j} = 2\pi (\cos{\theta} \ \hat{i} + \sin{\theta} \ \hat{j})$ $d\vec{r} = 2\pi (-\sin{\theta} \ \hat{i} + \cos{\theta} \ \hat{j}) \ d\theta$

$\vec{G} = (x+y) \ \hat{i} + \cos^2(\sqrt{x^2+y^2}) \ \hat{j} + x^2 \ \hat{k}$

The vector field evaluated at the boundary is:

$\vec{G} = 2\pi(\sin{\theta}+\cos{\theta}) \ \hat{i} + \hat{j} + 4\pi^2 \cos^2{\theta} \ \hat{k}$

$\vec{G} \cdot d\vec{r} = (-4\pi^2\sin{\theta}(\sin{\theta}+\cos{\theta})+2\pi\cos{\theta}) \ d\theta$

$\oint_{C} \vec{G} \cdot d\vec{r} =\int_{0}^{2\pi} (-4\pi^2\sin{\theta}(\sin{\theta}+\cos{\theta})+2\pi\cos{\theta}) \ d\theta = -4\pi^3$

$\implies\boxed{ \biggr\lvert \int \int_{S} \vec{F} \cdot d\vec{S} \biggr\rvert = 4\pi^3}$

This integral can also be solved by:

• Solving the surface integral without the use of any theorems. This means that the solver would have to evaluate the curl of the field and parameterise the boundary, which would involve more work.

• Closing the surface by a circle of radius of $2\pi$ and applying the divergence theorem to solve the integral. The resulting integral would be a volume integral. This also involves a fair amount of work.

I cannot think of other methods to solve this, but I have not used the additionally mentioned methods to cross-check my answer.

I used the following parametrization: $r=<x,y,\cos{\sqrt{x^2+y^2}}>.$ The integral is equivalent to: $\int_{0}^{2\pi}\int_{0}^{2\pi}\Big(\nabla \times G\Big) \sdot (r_x\times r_y)dxdy$ $=\int_{0}^{2\pi}\int_{0}^{2\pi}<-2\cos{r},2r\cos{\theta},-1>\sdot <\sin{r}\cos{\theta}, \sin{r}\sin{\theta}, 1>r dr d\theta$ $=\int_{0}^{2\pi}\int_{0}^{2\pi}\Big[-2\cos{r}\sin{r}\cos{\theta}+2r\sin{r}cos{\theta}\sin{\theta}-1\Big]r dr d\theta$ $=\int_{0}^{2\pi}\int_{0}^{2\pi} (-1)r dr d\theta$ $=-2\pi\Big[\frac{1}{2}r^2\Big]_{0}^{2\pi}$ $=-4\pi^3.$