Suppose that $C$ is the intersection of $z=2x+5y$ and $x^2+y^2=1$ which is oriented counterclockwise when viewed from above. Now let $$F=\langle \sin{x}+y, \sin{y}+z, \sin{z}+x \rangle$$
How can I find $\int_C{F\bullet dr}$?
So far I know that the intersection is between a plane and a cylinder. And I think that Stokes' Theorem should be usable in this problem.
We are given a vector $\vec F$ with
$$\vec F=\hat x (y+\sin x)+\hat y (z+\sin y)+\hat z (x+\sin z)$$
and asked to find the line integral $I$
$$I=\oint_C \vec F\cdot d\vec r$$
where the contour $C$ is formed by the intersection of the cylinder $x^2+y^2=1$ and the plane $z=2x+5y$.
We will use Stokes' Theorem to find $I$.
STOKES' THEOREM
Stokes' Theorem states that for sufficiently smooth vector fields $\vec F$ and $C$ a sufficiently smooth space curve, we have $$\oint_C \vec F\cdot d\vec r=\iint_S \nabla \times \vec F \cdot n dS$$ where $C$ is a contour that bounds the surface $S$.
First we find that the curl of $\vec F$ is $\nabla \times \vec F=-(\hat x+\hat y+ \hat z)$.
Next, we find that the unit normal to $S$. This is just the unit normal to the plane $z=2x+5y$. Therefore, $\hat n=\frac{-\hat x 2-\hat y5+\hat z}{\sqrt{30}}$ and $\vec F \cdot \hat n=\frac{6}{\sqrt{30}}$.
Finally, we need to find the element of surface area $dS$ of the surface of intersection between the plane and the cylinder. To do this, we observe that this area is the area of an element $dx\,dy$ divided by the cosine of the angle $\gamma$ between the $z$ axis and the plane. That is $\cos \gamma=\hat z \cdot \hat n=1/\sqrt{30}$ and $dS=\frac{dx\,dy}{\cos \gamma}$ .
Putting this all together gives
$$\begin{align} I&=\oint_C \vec F \cdot d\vec r\\\\ &=\iint_S \nabla \times \vec F \cdot \hat n\,dS\\\\ &=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\left(\frac{6}{\sqrt{30}}\right)\frac{dx\,dy}{\frac{1}{\sqrt{30}}}\\\\ &=\int_0^{2\pi}\int_0^1 \left(\frac{6}{30}\right)\frac{1}{\frac{1}{\sqrt{30}}}\rho\,d\rho\,d\phi\\\\ &=12 \pi \end{align}$$
