killjoy1990xb9

2021-12-30

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

limacarp4

Beginner2021-12-31Added 39 answers

(i) Given: $r=a(1-\mathrm{sin}\theta )$

$\frac{dr}{d\theta}=-a\mathrm{cos}\theta$

from (1)$a=\frac{r}{1-\mathrm{sin}\theta}$

$\Rightarrow \frac{dr}{d\theta}=\frac{-r\mathrm{cos}\theta}{1-\mathrm{sin}\theta}$

$\Rightarrow \frac{dr}{d\theta}+\frac{r\mathrm{cos}\theta}{1-\mathrm{sin}\theta}=0$

from (1)

hysgubwyri3

Beginner2022-01-01Added 43 answers

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

Vasquez

Expert2022-01-09Added 669 answers

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