areacalculusintegrationpolar coordinatestrigonometry
I think the formula is
$$A = \frac 1 2 \int_{\alpha}^{\beta} (\text{outer})^2 – (\text{inner})^2 d\theta$$
where $\alpha, \beta$ are where they intersect in $[0, 2\pi]$.
This is what I got based on that
By symmetry, we have
$$\frac A 4 = \frac 1 2 \int_{0}^{\pi/6} (8 cos 2\theta) – (2)^2 d\theta$$
Is that right?
You are calculating the area in the following graph
and your calculation is correct.
But the question asks for the area "inside" both graphs which can be seen below:
This area is then $$A=\int_0^{\pi/3}(1-\cos x)^2dx+\int_{\pi/3}^{\pi/2}\cos^2xdx$$ which is what the back of your book says.
Never mind. I got it. We set $r=0$ because those give the $\theta$'s where the small loop starts and ends. This can be seen by retracting the polar graph based on the Cartesian graph of
$$y = 1+2\cos(x)$$
The small loop begins at $2\pi/3$ and ends at $4\pi/3$.
Best Answer
Which one is in your book? $$\frac A4=\int_0^{\frac{\pi}6}\int_2^{\sqrt{8\cos(2\theta)}}rdrd\theta=\int_0^{\frac{\pi}6}\frac12(8\cos(2\theta)-2^2)d\theta=2\sin(\frac{\pi}3)-\frac{\pi}3=0.68$$