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?
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)