Line integral of an exact expression So I got this question for an assignment:
After I parametrized the path, I realized how tedious it would be to sub it in and solve it as a line integral, however I do notice that the expression is exact, so is there any short cuts to solve it without using the parametrized path?
 A: The integral is zero, and there are two "shortcuts" to get it provided that you are permitted to use the knowledge in the "shortcuts".
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Suppose vector filed $\,\mathbf F\,=\,y^2z^3\vec i\,+\,2xyz^3\vec j\,+\,3xy^2z^2\vec k$, then the integral becomes 
$$\int_{\gamma}\mathbf F\,{\rm d}\mathbf r$$
with $\,{\rm d}\mathbf r=dx\vec i+dy\vec j+dz\vec k\,$
Then, we calculate the curl of $\,\mathbf F$
$\rm curl(\mathbf F)\ =\ \left|\begin{matrix}\vec i&\vec j&\vec k\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ y^2z^3&2xyz^3&3xy^2z^2\end{matrix}\right|\ =\ 0$
$\left.\right.$
This shows $\,\mathbf F\,$ is a conservative vector field, so the line integral of $\,\mathbf F\,$ along a path only depends on the start point  and the endpoint of that path. In your question, the path is a closed loop, on which any start point will also be a endpoint, so the integral is taken on a single point and it has to be zero.
(Link of conservative field: https://en.wikipedia.org/wiki/Conservative_vector_field)
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Another way is to use Stoke's theorem:
$$\int\int_\Sigma\left(\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\,dz\,+\,\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\,dx\,+\,\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy\right)$$
$$=\oint_{\partial\Sigma}(\,P\,dx\,+\,Q\,dy\,+\,R\,dz\,)$$
where $\,\Sigma\,$ is a surface with its boundary $\,\partial\Sigma$ 
Apply the theorem in your question and we have
$$P=y^2z^3,\ \ Q=2xyz^3,\ \ R=3xy^2z^2$$
Work out the partial derivatives
$$\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=0$$
$$\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}=0$$
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$$
which leads to a zero integral
(Link of Stoke's Theorem:https://en.wikipedia.org/wiki/Stokes_theorem) 
