Question

The value of the line integral $\frac{2}{\pi }{\oint }_c(-y^{3}dx+x^{3}dy)$,
where C is the circles $x^2+y^2=1$ oriented counter - clockwise, is

• A. 0
• B. 1
• C. 3
• D. 6

Answer 1

The correct option is C 3

Using Green's theorem:

${\oint }_{c}Pdx+Qdy=\int \underset{R}{\int }(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy$

$\because \frac{2}{\pi }{\oint }_c(-y^{3}dx+x^{3}dy)$

Here P = $-y^3,Q=x^3$

So, we can write

$=\frac{2}{\pi }\int \underset{R}{\int }(\frac{\partial \left(x^3\right)}{\partial x}-\frac{\partial (-y^3)}{\partial y})dxdy$

$=\frac{2}{\pi }\int \underset{R}{\int }3(x^2+y^2)dxdy$

$=\frac{2\times 3}{\pi }\int \underset{R}{\int }(x^2+y^2)dxdy$

Let x = r cos $\theta$

y = r sin $\theta$

and dx dy = r dr d$\theta$

so,
$=\frac{6}{\pi }\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{1}{\int }}(r^{2}co{s}^2\theta +r^{2}si{n}^2\theta )rdrd\theta$

$=\frac{6}{\pi }\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{1}{\int }}{r}^{3}drd\theta$

$=\frac{6}{\pi }\underset{0}{\overset{2\pi }{\int }}{\left[\frac{r^4}{4}\right]}_0^{1}d\theta$

$=\frac{6}{\pi }\times \frac{1}{4}\times \left[\theta {]}_0^{2\pi }\right[\because$ the radius of circle , r = 1]

$=\frac{6}{\pi }\times \frac{1}{4}\times 2\pi$

= 3