Question

Value of the integral ${\oint }_c(xydy-y^{2}dx)$, where C is the square cut from the first quadrant by the lines x = 1 and y = 1 will be (use Green's theorem to change the line integral into double integral)

• A. $\frac{1}{2}$
• B. 1
• C. $\frac{3}{2}$
• D. $\frac{5}{3}$

Answer 1

The correct option is C $\frac{3}{2}$

By Green's theorem

${\int }_c(f_{1}dx+f_{2}dy)=\int {\int }_R(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y})dxdy$
or ${\oint }_c(Mdx+Ndy)=\int {\int }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$
${\oint }_c(-y^{2}dx+xydy)=\int {\int }_R\left[(y-(-2y\left)\right)\right]dxdy$
$={\int }_{x=0}^1{\int }_{y=0}^1\left(3y\right)dydx$
$={\int }_0^1{\left[\frac{3}{2}{y}^2\right]}_{y=0}^{1}dx$
$=\frac{3}{2}{\int }_0^{1}dx=\frac{3}{2}(1-0)=\frac{3}{2}$