Just another line integral

I can try; feel free to let me know if you need further clarifications.

As a mathematician, I'm used to phrase this stuff in terms of "forms," but in introductory maths and physics it is usually expressed in terms of vector fields. In the problem at hand, we are looking at the vector field $\vec{F}(x,y,x)= (yz\cos x, z\sin x +\sin z +\cos(y^3), y(\sin x +\cos z +1))$ .

Did you learn about "conservative" vector fields? These can be characterised in various equivalent ways; (a) The circulation is always 0, (b) the line integral is path-independent (depending only on the end points of the oriented curve), and (c) $curl(\vec{F})=\vec{0}$ (if the vector field is defined on all of $\mathbb{R}^3$ , as it is in our case). If a vector field is conservative, it has a "potential" $f(x,y,z)$ , that is , a scalar field such that $grad(f)=\nabla f=\vec{F}$ . In this case, if $C$ is a curve from $A$ to $B$ , we have a fundamental theorem for finding line integrals, analogous to the single-variable case: $\int_C \vec{F}\cdot d\vec{s}=f(B)-f(A)$ .

The vector field $\vec{F}$ in our example is not quite conservative (check the curl), but if we split off the simple field $(0,0,y)$ and write $\vec{F}(x,y,z)=\vec{G}(x,y,z)+(0,0,y)$ , then the field $\vec{G}(x,y,x)= (yz\cos x, z\sin x +\sin z +\cos(y^3), y(\sin x +\cos z))$ turns out to be conservative. You can find a potential $f$ for $\vec{G}$ by integrating each component function partially and "reconciling" the answers (as I call it in my classes). For example, for the first component you get $f(x,y,z)=\int yz\cos(x)dx$ $=yz\sin(x)+p(y,z)$ . If you proceed analogously for all three components, you end up with $f(x,y,z) =yz\sin x+y\sin z+ F(y)$ where $F'(y)=\cos(y^3)$ (try it yourself and feel free to ask if you have more questions). Now you can apply the fundamental theorem, $\int_C \vec{G}\cdot d\vec{s}=f(B)-f(A)=f(-1,0,\pi)-f(1,0,0)=0$ .

We can easily do the remaining line integral over the simple vector field $(0,0,y)$ directly, by using the given parameterisation: $\int_C ydz=\int_{0}^{\pi} \sin^3t dt=\frac{4}{3}$ .

As you see in Mr Hennings solution, you can also do the whole problem simply by using the parameterisation and "plowing through it," but I suggest that you familiarise yourself with the special techniques we have for conservative vector fields.

I hope this makes some sense; please let me know.