a) Let $D \subset \mathbb{R}^2$ be the unit disk and $\partial D$ be its positive oriented boundary. Evaluate the counter clockwise line integral:
$$\oint \limits_{\partial D} (x-y^3, x^3-y^2)\, {\rm d}(x, y)$$
b) Can you deduce if the function $f(x, y) =(x-y^3, x^3-y^2)$ is a conservative field using the above question?
Solution
We are making use of Green's thoerem for the first part.
a) $f$ is of the form $f(x, y)=(P, Q)$ where $P(x, y)=x-y^3$ and $Q(x, y)= x^3-y^2$. From Green's theorem we have successively:
$$\begin{aligned}\oint \limits_{\partial D} \left ( x-y^3, x^3-y^2 \right ) {\rm d}\left ( x, y \right ) &=\iint \limits_{D} \left ( \frac{\partial }{\partial x}Q(x, y) - \frac{\partial }{\partial y} P(x, y)\right )\, {\rm d}(x, y) \\
&= \iint \limits_{D} \left ( 3x^2 + 3y^2 \right )\, {\rm d}(x, y)\\
&= 3\int_{0}^{1}\int_{0}^{2\pi} \, {\rm d\theta}\, {\rm d} \rho\\
&= 6\pi \neq 0
\end{aligned}$$
b) If $f$ was a conservative field then the line integral over all closed curves would have to be $0$. However in question (a) we have found a closed curve that the line integral is not $0$. Hence $f$ is not a conservative field.
$$\oint \limits_{\partial D} (x-y^3, x^3-y^2)\, {\rm d}(x, y)$$
b) Can you deduce if the function $f(x, y) =(x-y^3, x^3-y^2)$ is a conservative field using the above question?
Solution
We are making use of Green's thoerem for the first part.
a) $f$ is of the form $f(x, y)=(P, Q)$ where $P(x, y)=x-y^3$ and $Q(x, y)= x^3-y^2$. From Green's theorem we have successively:
$$\begin{aligned}\oint \limits_{\partial D} \left ( x-y^3, x^3-y^2 \right ) {\rm d}\left ( x, y \right ) &=\iint \limits_{D} \left ( \frac{\partial }{\partial x}Q(x, y) - \frac{\partial }{\partial y} P(x, y)\right )\, {\rm d}(x, y) \\
&= \iint \limits_{D} \left ( 3x^2 + 3y^2 \right )\, {\rm d}(x, y)\\
&= 3\int_{0}^{1}\int_{0}^{2\pi} \, {\rm d\theta}\, {\rm d} \rho\\
&= 6\pi \neq 0
\end{aligned}$$
b) If $f$ was a conservative field then the line integral over all closed curves would have to be $0$. However in question (a) we have found a closed curve that the line integral is not $0$. Hence $f$ is not a conservative field.
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