Parametrization of $y^2=x^3$ to solve a line integral?

Question

Evaluate the following vector line integral given in differential form $$\int_{\gamma} x^2y\,dx-xy\,dy$$ where $\gamma$ is the curve with equation $y^2=x^3$, from $(1,-1)$ to $(1,1)$. The textbook also says that a parametrization of $\gamma$ is $$\vec{r}(t)=(t^2,t^3),\ -1\le t\le 1$$ My question is: How do I find these parametrizations? I know how to parametrize lines in plane and space, but what about a curve such as $y^2=x^3$? Should I give $x$ any value depending on $t$ and then find $y$? For instance, $$x=t \rightarrow y=t^{3/2}$$ What about the interval in which $t$ varies?

Answer 1

A rough sketch of the curve between start and end points may help.

As you can see, both start and end points $A(1,-1)$ and $B(1,1)$ have $x = 1$ but there are really two parts to the curve, one below x-axis and one above. A parametrization of $(t, t^{3/2})$ will only represent the curve above x-axis and you will need $(t, - t^{3/2})$ to cover the curve below x-axis. This will also mean, you will have to split your line integral in two parts.

i) $(t, - t^{3/2}), 0 \leq t \leq 1$ and your integral will have lower bound of $t=1$ and upper bound of $t = 0$.

ii) $(t, t^{3/2}), 0 \leq t \leq 1$ and your integral will have lower bound of $t=0$ and upper bound of $t = 1$.

That will ensure you line integral is along the curve from $A(1,-1)$ to $B(1,1)$.

But splitting a line integral for the same curve is not ideal and the parametrization $(t^2, t^3)$ addresses this issue because it represents your entire curve from $A$ to $B$ for $-1 \leq t \leq 1$.