$\displaystyle R=a(1+cos\theta{})\ is\ a\ cardiode. \\[2ex] \displaystyle F=-xy(xi-yj)\ \\[2ex] F=\ -x^2yi+xy^2j\ \\[2ex] dr=dxi+dyj+dzk \\[2ex] \displaystyle \therefore{}F\bullet{}dr=(-x^2yi+xy^2j)\bullet{}(dxi+dyj+dzk) \\[2ex] \displaystyle F\bullet{}dr=-x^2ydx+xy^2dy….\left(1\right) \\[2ex] $

$\displaystyle Let\ P=-x^2\ and\ Q=xy^2 \\[2ex] \displaystyle \therefore{}\frac{\partial{}P}{\partial{}y}=-x^2\ and\ \frac{\partial{}Q}{\partial{}x}=\ y^2 \\[2ex] \displaystyle By\ Green’s\ Theorem,\\[2ex] \displaystyle \int_cPdx+Qdy=\iint_R\left(\frac{\partial{}Q}{\partial{}x}-\frac{\partial{}P}{\partial{}y}\right)dx\ dy \\[2ex] \displaystyle \int_c-x^2ydx+xy^2dy=F.dr\ \iint_R\left(y^2+x^2\right)dx\ dy…(2)\\[4ex] From\ 1\ and\ 2,\ \\[3ex] \displaystyle \int_cF.dr=\ \iint_R\left(y^2+x^2\right)dx\ dy\\[3ex] $

$\displaystyle Converting\ to\ polar\ form, \\[2ex] \displaystyle X=r\cos\theta{};\ y=r\sin\theta{};\ dxdy=rdrd\theta{} \\[2ex] $

$\displaystyle For\ complete\ cardiode,\\[2ex] limits\ of\ \theta{}\ are\ 0\ to\ 2\pi{}\ and\ limits\ of\ r\ are\ 0\ to\ a(1+cos\theta{}) \\[2ex] $

$\begin{align*} \displaystyle \int F.dr&=\int_0^{2\pi{}\ }\int_0^{a(1+\cos\theta{})}\left({r^2\cos}^2\theta{}+{r^2\sin}^2\theta{}\right)\bullet{}rdrd\theta{} \\[2ex] \displaystyle &=\ \int_0^{2\pi{}\ }\int_0^{a(1+\cos\theta{})}\left({\cos}^2\theta{}+{\sin}^2\theta{}\right)\bullet{}r^3drd\theta{} \\[2ex] \displaystyle &=\int_0^{2\pi{}\ }{\left[\frac{r^4}{4}\right]}_0^{a(1+\cos\theta{})}d\theta{}\\[2ex] &=\frac{1}{4}\int_0^{2\pi{}}{{[a}^4(1+\cos\theta{})}^4-0]d\theta{} \\[2ex] \displaystyle &=\frac{a^4}{4}\int_0^{2\pi{}}{\left[{2\cos}^2\left(\frac{\theta{}}{2}\right)\right]}^4d\theta{}\\[2ex] &=\frac{a^4}{4}\times{}2^4\int_0^{2\pi{}}\left[{\cos}^8\left(\frac{\theta{}}{2}\right)d\theta{}\right] \\[2ex] \end{align*} $

$\displaystyle Put\ \theta{}/2=t, \\[2ex] \displaystyle \therefore{}\theta{}=2t \\[2ex] \displaystyle \therefore{}d\theta{}=2dt;\ \\[2ex] $

| $\theta$ | $0$ | $2\pi$ | | --- | --- | --- | | $t$ | $0$ | $\pi$ |  $\displaystyle \therefore{}\int F\bullet{}dr=4a^4\int_0^{\pi{}\ }{cos}^8t\bullet{}2dt \[4ex]

\displaystyle Using\ theorem, \[3ex]

\displaystyle \therefore{}\ \int_0^{2a}f\left(t\right)dt=2\int_0^af\left(t\right)dt\ \ \ \ \ \ \ \ \left[when\ f\left(2a-t\right)=f\left(t\right)\right]\ \ \[2ex]

\displaystyle \therefore{}\ \int_0^{2a}f\left(t\right)dt=0\ \ \ \ \ \ \ \ \ [when\ f(2a-t)=-f(t)] \[2ex] $ $\displaystyle \therefore{}\int F\bullet{}dr=4a^4\bullet{}2\bullet{}2\int_0^{\pi{}/2}{cos}^8tdt \[2ex]

\displaystyle =16a^4\times{}\frac{7}{8}\times{}\frac{5}{6}\times{}\frac{3}{4}\times{}\frac{1}{2}\times{}\frac{\pi{}}{2}\ \left(reduction\ Formula\right) \[2ex]

\displaystyle =35\pi{}\frac{a^4}{16} \[2ex]

$