areaintegrationpolar coordinates
I need to find the area between two polar curves, $$r = \frac1{\sqrt{2}}$$ $$r = \sqrt{\cos(θ)}$$
I've found the intersections to be at $\fracπ3$ and $\frac{5π}3$, and I've set up the equation to find the area as
$$\int\limits_{\fracπ3}^{\frac{5π}3} \sqrt{\cos(θ)}^2 – \frac1{\sqrt{2}}^2 \, \mathrm dθ,$$
but whenever I plug it into a calculator, it comes up as undefined, so it can't possibly be correct. Could you help me with this?
A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.
The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$ \underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region} + \underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1. $$
In this case it is more convenient to consider
\begin{align} f_1(y)&=3-\tfrac14\,y^2 ,\\ f_2(y)&=y . \end{align}
\begin{align} S&= \int_{-6}^2\!\int_y^{3-\tfrac14\,y^2}\!dx\,dy = \left. 3y-\tfrac1{12}y^3-\tfrac12 y^2\right|_{-6}^2 =\tfrac{64}3 . \end{align}
Best Answer
The area show in green
is given by the following integral, where $r(\theta)=\sqrt{\cos\theta}$ $$ \int_{\pi/3}^{\pi/2}d\theta\int_0^{r(\theta)}rdr=\frac{1}{2}\int_{\pi/3}^{\pi/2}\cos\theta d\theta=\frac{1}{2}\left(1-\frac{\sqrt{3}}{2}\right) $$
The rest should be pretty easy.