killjoy1990xb9

2021-12-30

Form the differential equations by eliminating arbitrary constant a. $r=a\left(1-\mathrm{sin}\theta \right)$

limacarp4

(i) Given: $r=a\left(1-\mathrm{sin}\theta \right)$
$\frac{dr}{d\theta }=-a\mathrm{cos}\theta$
from (1) $a=\frac{r}{1-\mathrm{sin}\theta }$
$⇒\frac{dr}{d\theta }=\frac{-r\mathrm{cos}\theta }{1-\mathrm{sin}\theta }$
$⇒\frac{dr}{d\theta }+\frac{r\mathrm{cos}\theta }{1-\mathrm{sin}\theta }=0$

hysgubwyri3

Now obviously, our area is equal to $\frac{\pi }{2}{a}^{2}+2{\int }_{0}^{\frac{\pi }{2}}\frac{1}{2}{\left[a\left(1-\mathrm{sin}\left(\theta \right)\right)\right]}^{2}d\theta$

Vasquez

$\begin{array}{}r=a\left(1-\mathrm{sin}\theta \right)\\ a\left(1-\mathrm{sin}\left(\theta \right)\right)=r\\ a-a\mathrm{sin}\left(\theta \right)=r\\ \left(1-\mathrm{sin}\left(\theta \right)\right)a=r\\ \left(-\mathrm{sin}\left(\theta \right)+1\right)a=r\\ \frac{\left(-\mathrm{sin}\left(\theta \right)+1\right)a}{-\mathrm{sin}\left(\theta \right)+1}=\frac{r}{-\mathrm{sin}\left(\theta \right)+1}\\ a=\frac{r}{-\mathrm{sin}\left(\theta \right)+1}\end{array}$

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