Answer to Question #151185 in Calculus for Zeeshan

Question

∫ x² y dx + x y² dy
where d is describe by 0 ≤ x ≤ 1 , 0 ≤ y ≤ x

Answer 1

Consider the region $D$ as shown in the figure below:

The region $D$ is a closed smooth curve, so the Green's theorem applicable for this region.

Using Green's theorem, the line integral is evaluated as,

$\int x^2ydx+xy^2dy$

$=\iint_D(\frac{\partial}{\partial x}(xy^2)-\frac{\partial}{\partial y}(x^2y))dA$

$=\iint_D(y^2-x^2)dA$

$=\int_{x=0}^{1}\int_{y=0}^{x}(y^2-x^2)dydx$

$=\int_{x=0}^{1}[\frac{y^3}{3}-x^2y]_{y=0}^{x}dx$

$=\int_{x=0}^{1}(\frac{x^3}{3}-x^3)dx$

$=-\frac{2}{3}\int_{x=0}^{1}x^3dx$

$=-\frac{2}{3}[\frac{x^4}{4}]_{x=0}^{1}$

$=-\frac{2}{3}(\frac{1}{4})$

$=-\frac{1}{6}$

Therefore, the line integral is $\int x^2ydx+xy^2dy=-\frac{1}{6}$