Integral of a Gradient function (and another function?) I'm aware of line integrals around planes and curves, but I cannot make up how to approach this question:
$$\int_C f∇f \cdot \,d\mathbf{r} $$ where $f(x,y,z)=xz\cos(x^2+y^2)$ and C is the intersection of the cylinder $x^2+y^2=1$ and $x+y+2z=2$
Suppose I find the intersection, does the integral constitute of a product of a gradient function and its function? If so, how do I approach this?
 A: Note from the product rule for differentiation, we can write 
$$f(\vec r)\nabla f(\vec r)=\frac12 \nabla (f^2(\vec r))$$
In addition, the line integral of the gradient of a smooth function $\phi$ can be simplified as 
$$\begin{align}
\int_C  \nabla \phi(\vec r) \cdot \,d\vec r&=\int_{t_1}^{t_2} \nabla \phi(\vec r(t)) \cdot \frac{d\vec r}{dt}\,dt\\\\
&=\int_{\phi_1}^{\phi_2} d\phi \\\\
&=\phi(\vec r_2)-\phi(\vec r_1) \tag 1
\end{align}$$
And if $C$ is closed, then $\vec r_2=\vec r_1$ and the line integral is zero.  So, for any (sufficiently smooth) closed curve $C$, $(1)$ becomes
$$\oint_C \nabla \phi(\vec r)\cdot d\vec r=0 \tag 2$$
The intersection of the cylinder $x^2+y^2=1$ with the plane $x+y+2z=2$ is indeed a closed curve which can be parametrically described as
$$\begin{align}
x&=\cos(t)\\\\
y&=\sin(t)\\\\
z&=1-\frac12 (\cos(t)+\sin(t))
\end{align}$$
for $t\in [0,2\pi]$.
Denote this closed curve by $C$ and let $f^2=\phi$ in $(2)$.  Thus, we have
$$\oint_C f(\vec r)\nabla f(\vec r)\cdot \,d\vec r=0$$
