# Find the circulation around a closed contour which is defined by $$x = 1, y = 0, y = 1$$ and $$x = 0$$.

A velocity field is given by $$u=x^{2} + y$$ and $$v= x^{3} + 2y$$. Find the circulation around a closed contour which is defined by $$x = 1, y = 0, y = 1$$ and $$x = 0$$. Units of $$u$$ and $$v$$ are in $$m/s$$.

Asked on 15th April 2021 in

Circulation around a rectangular contour

The closed contour is a square and its corner are at $$A(0,0), B(1,0), C(1,1), D(0,1)$$.
Circulation is given by $\Gamma =\int _{ABCD}udx + vdy$
$\Rightarrow\Gamma = \int_{A}^{B}udx + \int_{B}^{C}vdy+\int_{C}^{D}udx+\int_{D}^{A}vdy$
$=\int_{0}^{1}\left ( x^{2}+y \right )dx + \int_{0}^{1}\left ( x^{3}+2y \right )dy+\int_{1}^{0}\left ( x^{2}+ y \right )dx + \int_{1}^{0}\left ( x^{3}+2y \right )dy$ In the first integral, $$y=0$$, in second integral $$x=1$$, in the third integral, $$y=1$$ and in the fourth integral $$x=0$$.
$=\int_{0}^{1}\left ( x^{2} \right )dx + \int_{0}^{1}\left ( 1 + 2y \right )dy + \int_{1}^{0}\left ( x^{2} + 1 \right )dx + \int_{1}^{0}\left ( 2y \right ) dy$
$=\left | \frac{x^{3}}{3} \right |_{0}^{1}+\left | y+ \frac{2y^{2}}{2} \right |_{0}^{1}+\left | \frac{x^{3}}{3} + x \right |_{1}^{0}+\left | \frac{2y^{2}}{2} \right |_{1}^{0}$
$=\left [ \frac{1}{3} \right ]+\left [ 1+1 \right ]+\left [ -\frac{1}{3}-1 \right ]+\left [ -1 \right ]=0$

Answered on 30th April 2021.