1. ## [SOLVED] Stokes' Theorem

Evaluate $\oint_{C}(y^3+cosx)dx+(siny+z^2)dy+xdz$, where C is the closed curve parameterized and oriented by the path $\vec{x}(t)=(cost,sint,sin2t)$.

Hint: Show that the path $\vec{x}(t)$ lies on the surface $z=2xy$.

2. Originally Posted by davesface
Evaluate $\oint_{C}(y^3+cosx)dx+(siny+z^2)dy+xdz$, where C is the closed curve parameterized and oriented by the path $\vec{x}(t)=(cost,sint,sin2t)$.
"Parameterized", yes, but this doesn't give an orientation. Are we to assume that t is going from 0 to $2\pi$? That would give an orientation.

[/quote]Hint: Show that the path $\vec{x}(t)$ lies on the surface $z=2xy$.[/QUOTE]
Have you done the "hint"? Do you know any identity for "sin(2t)"?

Stokes' theorem, since you title this thread "Stokes' Theorem", in its simplest form says that $\vec{f}(x,y,z)$ is a vector valued function then $\oint_C \vec{f}\cdot d\vec{\sigma}= \int\int \nabla\times d\vec{S}$.

Here, $\vec{f}(x,y,z)= (y^3+ cos(x))\vec{i}+ (sin(y)+ z^2)\vec{j}+ x\vec{k}$ and $d\vec{\sigma}$ is the differential tangent vector, $(-sin(t)\vec{i}+ cos(t)\vec{j}+ 2cos(2t)\vec{k})dt$.

Since you know that the path lies entirely in the plane z= 2xy, you can write that as the position vector $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ 2xy\vec{k}$. The two derivative vectors, $\vec{r}_x(x,y)= \vec{i}+ 2y\vec{k}$ and $\vec{r}_y(x,y)= \vec{j}+ 2x\vec{i}$ are in the tangent plane to the surface (which here is the plane itself) and their cross product, $\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 2y \\ 0 & 1 & 2x\end{array}\right|= -2y\vec{i}- 2x\vec{j}+ \vec{k}$ is normal to the plane and its length gives the "differential of surface area". Assuming that the path integral was to be from 0 to $2\pi$ (rather than from 0 to $-2\pi$), the normal is to be oriented by positive z so $d\vec{S}= (-2y\vec{i}- 2\vec{j}+ \vec{k})dy dx$.

You still need to determine the bounds on x and y.

3. I don't know what to say about the orientation. I've posted all the information given about the problem. I'm guessing it's the negative orientation, though, since the given answer is $\frac{-3\pi}{4}$.

I have no clue how they got that, as my work has boiled down to $\int\int_{S} (\nabla \times F) \cdot \vec{N} dxdy= \int \int_{D}(8x^2+2x-3y^2)dxdy$ by taking the cross product and dot product where indicated and then substituting in 2xy for the z value that showed up from the dot product.

What am I missing here?

4. Wow, you know what? It's way easier than I made it seem. All you have to do is convert it to polar and reverse the sign on the integral. r goes from 0 to 1 and theta goes from 0 to 2pi. Thanks for the help.