I have a following exercise, falling under the topic of line integrals.
Calculate the integral: $$I=\int\limits_{\gamma} \sin(yz)\,dx+xz\cos(yz)\,dy+xy\cos(yz)\,dz$$ Where: $$\gamma(t)=(\cos^{4}t,\cos t\sin^5t,\sin^4t)$$ $$t \in[0,12\pi]$$ I based my solution on a chapter devoted to "Exact differential", so the task there was a little bit different. I need to calculate the integral and checking for path independence is (as I have read) an easier approach to solve such problems.
My solution:
I check if $$P'_{y}=Q'_{x}$$ $$Q'_{z}=R'_{y}$$ $$R'_{x}=P'_{z}$$ which would demonstrate that the integral is path independent.
And indeed, after calculating those partial derivatives, I obtain:
$$P'_{y}=z\cos(yz)$$ $$Q'_{x}=z\cos(yz)$$ $$Q'_{z}=x\cos(yz)-xyz\sin(yz)$$ $$R'_{y}=x\cos(yz)-xyz\sin(yz)$$ $$R'_{x}=y\cos(yz)$$ $$P'_{z}=y\cos(yz)$$
I write three conditions:
$$(1) \quad U'_{x}=\sin(yz)$$ $$(2) \quad U'_{y}=xz\cos(yz)$$ $$(3) \quad U'_{z}=xy\cos(yz)$$
From the $(1)$: $$U(x,y,z)=\int \sin(yz)\,dx=x\sin(yz) + \varphi(y,z)$$ $$U_{y}'=xz\cos(yz)+\varphi'_{y}(y,z)$$ $$Q=U'_{y}$$ $$xz\cos(yz)=xz\cos(yz)+\varphi'_{y}(y,z)$$ $$\varphi'_{y}(y,z)=0$$ Unfortunately it does not look like a correct result. I presume, that $U(x,y,z)=xz\cos(yz)$ is not the integral of the above example.
I have the following questions:
1) What for do I need the information that $$\gamma(t)=(\cos^{4}t,\cos t\sin^5t,\sin^4t)$$ $$t \in[0,12\pi]$$
2) Is my solution correct so far? If yes, what to do next?
Your check has reveiled that the given vector field ${\bf v}$ is exact, implying that there is a potential function $U:{\mathbb R}^3\to{\mathbb R}$ with $$\nabla U={\bf v}\ .\tag{1}$$ It s a basic theorem of vector analysis that in this case the line integral in question is given by $$U\bigl(\gamma(12\pi)\bigr)-U\bigl(\gamma(0)\bigr)=U(1,0,0)-U(1,0,0)=0\ ,$$ and this solves the problem at hand.
Since the given $\gamma$ is a closed curve it was not necessary to compute $U$ explicitly. For an arbitrary $\gamma$ we need the potential $U$ explicitly. It can be obtained in the following way: Writing out $(1)$ in coordinates we have $$U_x=\sin(yz),\quad U_y=xz\cos(yz),\quad U_z=xy\cos(yz)\ .$$ From the first of these equations we deduce that $$U(x,y,z)=x\sin(yz)+f(y,z)$$ with an unknown function $(y,z)\mapsto f(y,z)$. This implies that $$xz\cos(yz)+f_y(y,z)=U_y=xz\cos(yz)\ ,$$ from which we deduce that $f_y(y,z)\equiv0$, or $f(y,z)=g(z)$ for some function $g$ of the single variable $z$. This implies $$xy\cos(yz)+g'(z)=U_z=xy\cos(yz)\ ,$$ and therefore that $g$ is constant, which we may choose to be $=0$. It follows that $$U(x,y,z):=x\sin(yz)$$ should do the trick, and a quick check shows that this $U$ indeed satisfies $(1)$.
Given any $$\gamma:\quad t\mapsto{\bf r}(t):=\bigl(x(t),y(t),z(t)\bigr)\qquad(0\leq t\leq1)$$ we then have $$\int_\gamma{\bf v}\cdot d{\bf r}=U\bigl({\bf r}(1)\bigr)-U\bigr({\bf r}(0)\bigr)\ .$$