Calculus problem #1938

Let $P(x,y) = x^4$ and $Q(x,y) = xy$ . By Green’s theorem, we have $\displaystyle \oint_C (P(x,y)\ dx + Q(x,y)\ dy) = \iint_A \left(\frac{\partial Q(x,y)}{\partial x} - \frac{\partial P(x,y)}{\partial y}\right)$ , where $C$ is the path and $A$ is the area enclosed by the path. So, we have $\frac{\partial Q(x,y)}{\partial x} = y$ , $\frac{\partial P(x,y)}{\partial y} =0$ and $\frac{\partial Q(x,y)}{\partial x} - \frac{\partial P(x,y)}{\partial y} = y-0= y$ . Substituting this in to the double integral, we have

$\begin{aligned} \displaystyle \oint_C (x^4\ dx + xy\ dy) &= \int_0^1\int_0^{1-x} y\ dy\ dx \\ &= \int_0^1 \left[\frac{y^2}{2}\right]_{y=0}^{y=1-x}\ dx \\ &= \int_0^1 \left(\frac{1-2x+x^2}{2}\right)\ dx \\ &= \left[\frac{x}{2} - \frac{x^2}{2} + \frac{x^3}{6} \right]_0^1 \\ &= \frac{1}{6} \approx .16666\dots \end{aligned}$