Answer to Question #176680 in Calculus for Joshua

Question

Find the area inside the inner loop of r=3-8cos∅.

Answer 1

Given curve  is symmetric with respect to the $x$ -axis. Then consider lower part of the figure bounded by the inner loop.

General equation of given curve is

$r=l-a\cos\theta$

The basic formula for area in polar coordinates is

$\displaystyle S=\frac{1}{2}\int\limits_\alpha^\beta r^2d\theta$

For the lower part of the inner loop

$\alpha=0$

$\beta=\arccos\left(\dfrac{l}{a}\right)$

To answer the question take the area twice

$\displaystyle S=2\cdot\frac{1}{2}\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(3-8\cos\theta\right)^2d\theta=$

$\displaystyle\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(3^2-2\cdot3\cdot8\cos\theta+(8\cos\theta)^2\right)d\theta=$

$\displaystyle\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(9-48\cos\theta+64\cos^2\theta\right)d\theta=$

use formula $\cos^2\theta=\dfrac{1}{2}(1+\cos2\theta)$

$\displaystyle\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(9-48\cos\theta+64\cdot\frac{1}{2}(1+\cos2\theta)\right)d\theta=$

$\displaystyle\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(9-48\cos\theta+32(1+\cos2\theta)\right)d\theta=$

$\displaystyle\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(9-48\cos\theta+32+32\cos2\theta\right)d\theta=$

$\displaystyle\int\limits_0^{\arccos\left(\frac{3}{8}\right)}\left(41-48\cos\theta+32\cos2\theta\right)d\theta=$

take the antiderivative

$\displaystyle\left[41\theta-48\sin\theta+32\cdot\frac{1}{2}\sin2\theta\right]_0^{\arccos\left(\frac{3}{8}\right)}=$

$\displaystyle\Bigg[41\theta-48\sin\theta+16\sin2\theta\Bigg]_0^{\arccos\left(\frac{3}{8}\right)}=$

$\displaystyle\Bigg[41\arccos\left(\frac{3}{8}\right)-48\sin\left(\arccos\left(\frac{3}{8}\right)\right)+16\sin\left(2\cdot\arccos\left(\frac{3}{8}\right)\right)-\Big(41\cdot0-48\sin0+16\sin(2\cdot0)\Big)\Bigg]=$

simplify

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sin\left(\arccos\left(\frac{3}{8}\right)\right)+16\sin\left(2\cdot\arccos\left(\frac{3}{8}\right)\right)=$

use $\sin\alpha=\sqrt{1-\cos^2\alpha}$ when $0<\alpha<\dfrac{\pi}{2}$

and $\sin(2\alpha)=2\sin\alpha\cos\alpha$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sqrt{1-\cos^2\left(\arccos\left(\frac{3}{8}\right)\right)}+16\cdot2\cdot\sin\left(\arccos\left(\frac{3}{8}\right)\right)\cdot\cos\left(\arccos\left(\frac{3}{8}\right)\right)=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sqrt{1-\left(\frac{3}{8}\right)^2}+32\cdot\sin\left(\arccos\left(\frac{3}{8}\right)\right)\cdot\frac{3}{8}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sqrt{1-\left(\frac{3}{8}\right)^2}+32\cdot\sqrt{1-\left(\frac{3}{8}\right)^2}\cdot\frac{3}{8}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sqrt{1-\frac{9}{64}}+32\cdot\sqrt{1-\frac{9}{64}}\cdot\frac{3}{8}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sqrt{\frac{64-9}{64}}+32\cdot\sqrt{\frac{64-9}{64}}\cdot\frac{3}{8}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\sqrt{\frac{55}{64}}+32\cdot\sqrt{\frac{55}{64}}\cdot\frac{3}{8}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-48\frac{\sqrt{55}}{8}+32\cdot\frac{\sqrt{55}}{8}\cdot\frac{3}{8}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-6\sqrt{55}+\frac{3\sqrt{55}}{2}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)+\frac{3}{2}\sqrt{55}-6\sqrt{55}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)+\left(\frac{3}{2}-6\right)\sqrt{55}=$

$\displaystyle41\arccos\left(\frac{3}{8}\right)-\frac{9}{2}\sqrt{55}$