How do I apply Stokes' theorem on this differential form? I want to calculate the integral
$ \int_\gamma \log(1+x^2)dx + \cos(1+y^2)dy +  (\sin(x^4) + y + \sin(z^2))dz$
where $\gamma = (\cos(t) + \sin(t), \cos(t)+ 2\sin(t), \cos(t) - \sin(t))$ for $0 \leq t \lt 2\pi$.
I want to apply Stoke's theorem, and so I've calculated
$dw  = (\cos(x^4 4x^3) dx \wedge dz +  dy \wedge dz$.
I don't know how to integrate this over the area inside $\gamma$.
Any help would be appreciated!
 A: Notice that projecting the curve onto the $xz$-plane gives a circle of radius $\sqrt{2}$, since
$$ x^2 + z^2 = (\cos(t)+\sin(t))^2 + (\cos(t)-\sin(t))^2 = 2\cos^2(t) + 2\sin^2(t) = 2 $$
So it will be easiest to integrate over the area's "shadow" in the $xz$-plane. Also notice that the curve lies entirely on the plane $3x - 2y - z = 0$ (also easily verified by plugging in the components of $\gamma$ for $x,y,z$), so we can write $y$ as a function of $x$ and $z$ as
$$ y = \frac{3x-z}{2} $$
Therefore, by taking the differential, we get
$$ dy = \frac{3dx - dz}{2} $$
Let's denote by $R$ the region in the $xz$-plane which is the projection/shadow of the area we are integrating over. Then the integral becomes
$$
\begin {eqnarray}
\int dw &=& \int_R 4x^3 \cos(x^4) \, dx \wedge dz + \frac{1}{2}\int_R (3dx - dz) \wedge dz \\
&=& \int_R 4x^3 \cos(x^4) \, dx \wedge dz + \frac{3}{2} \int_R \, dx \wedge dz
\end {eqnarray}
$$
The second integral is just $3/2$ times the area of the region $R$. Since $R$ is a disc of radius $\sqrt{2}$, this area is $2\pi$, and so the second integral is $3\pi$.
The integrand of the left/first integral depends only on $x$, and is an odd function. Since the region $R$ is symmetric about the $z$ axis, the integral will be zero. So overall, I think the entire integral will be $3\pi$.
