I have attempted to solve a polar curves question and I keep getting the wrong answer.
The question is:
Find the area enclosed by between the curves
$$
r = 3 – 3\cos\theta
$$
and
$$
r =4\cos\theta
$$
So far I have forked out the angle that they interest at so:
$$
4cos\theta = 3 – 3cos\theta
$$
$$
\therefore \theta = 1.127885 radians
$$
Then I integrating with the upper bound as 1.127885 and the lower bound as zero.
$$
\int_{0}^{1.127885}{\frac{1}{2}*(4\cos\theta)^2}d\theta = 6.06037
$$
$$
\int_{0}^{1.127885}{\frac{1}{2}*(3-3\cos\theta)^2} d\theta = 0.3527
$$
Then the difference in area is 6.06037 – 0.3527 = 5.70767.
Then by symmetry the total area is 5.70567*2 = 11.4
The answer I get is 11.4 but apparently the answer 1.15. Could you help me with this?
Best Answer
The intersection angle is given by $\cos\theta =\frac37$ and the integral for the enclosed area is given by
$$\begin{align} &2 \int_0^{\arccos\frac37} \frac12(3-3\cos \theta)^2 d\theta + 2\int_{\arccos\frac37}^{\pi/2}\frac12 (4\cos\theta)^2d\theta \\ =& \>4\pi -\frac{39\sqrt{10}}7+\frac{11}2\arccos\frac37=1.151 \end{align}$$
Note that the limits in the two integrals above ensure that the area is enclosed inside both curves.