Line integral of $e^{x^3} \vec{i} + y^3 \vec{j} + e^{z^2} \vec{k}$

Question

I need to find:

$\oint_C (e^{x^3} \vec{i} + y^3 \vec{j} + e^{z^2} \vec{k})\cdot d\vec{r}$

where $C$ is the intersection between the surfaces $z=x^2+y^2$ and $z=x-3y$.

I know I need to parametrize $C$, but I'm not sure how to do so.

Answer 1

We can find $C$ as follows: $$x^2 + y^2 = x-3y \\ \implies x^2 -x + y^2+3y =0$$ Which, upon completing the square, gives us $$\displaystyle \bigg(x- \frac{1}{2}\bigg)^2 + \bigg(y+\frac{3}{2}\bigg)^2=\frac{5}{2}$$ Which is a circle with midpoint $(\frac{1}{2},-\frac{3}{2})$ and radius $\sqrt{\frac{5}{2}}$.

This parameterises as follows \begin{align*}x &= \frac{1}{2}+\sqrt{\frac{5}{2}}\cos{\theta} \\ y &= -\frac{3}{2} + \sqrt{\frac{5}{2}}\sin{\theta} \\ 0&\leq\theta \leq2\pi\end{align*}

Answer 2

Equating the two expressions gives

$$x^2+y^2 = x-3y.$$

This is a displaced circumference. Complete squares to find the radius and then apply the usual parameterization.