Parametrization

Question

How do I parametrize the attached problems. I am a bit confused because it is not in a typical format that can be easily parametrized.

Answer 1

I'll work out one example to get you started; I'll take (d).

The region $\mathcal{R}$ is the triangle connecting $A=(0,0)$, $B=(1,0)$ and $C=(1,1)$. You need to parametrize the three sides; I'll choose parametrizations in such a way that the parameter $t$ runs from $0$ to $1$ in all cases:

• the segment connecting $A$ and $B$:$$\vec r_1 = (t,0)$$
• the segment connecting $B$ and $C$:$$\vec r_2 = (1,t)$$
• the segment connecting $C$ and $A$:$$\vec r_3 = (1-t,1-t)$$

Then: $$\begin{align} \oint_{\mathcal{C}}xy^2dx+x^2ydy & =\int_{{AB}}xy^2dx+x^2ydy+\int_{{BC}}xy^2dx+x^2ydy+\int_{{CA}}xy^2dx+x^2ydy \\[8pt] & =\int_0^1 t0^2dt+t^20d0+\int_0^1t^2d1+tdt+\int_0^1(1-t)^3d(1-t)+(1-t)^3d(1-t) \\[8pt] & =\int_0^1 \left( t -2(1-t)^3\right) dt\\[8pt] & = \ldots \\[8pt] & =0 \end{align}$$

This is easily verified with Green's theorem since: $$\frac{\partial (x^2y)}{\partial x}-\frac{\partial (xy^2)}{\partial y} = 2xy-2xy = 0$$ So clearly: $$\oint_{\mathcal{C}}xy^2dx+x^2ydy = \iint_{\mathcal{R}} \frac{\partial (x^2y)}{\partial x}-\frac{\partial (xy^2)}{\partial y} \, dydx = \int_0^1 \int_0^x 0 \, dydx = 0$$