Find the common area between the two polar functions: $r=\sin\theta$ and $r=\cos\theta$; however, I'm not too sure about how to set up the bounds for the integral. I know the functions have the same value at $\theta=\pi/4$ and $\theta=5\pi/4$. If so, would the integral look something like this?
$$A=\int_{0}^{\pi/2}\left( \sin^2\theta + \cos^2\theta\right)d\theta$$
On
Black curve: $r = \sin \theta$
Blue curve: $r = \cos \theta$
Shaded region is lamina of interest.
First, express the parametric curves in terms of Cartesian coordinates: $$ \begin{align} % \color{blue}{x^2+\left(y-\frac{1}{2}\right)^2} &\color{blue}{=\left(\frac{1}{2}\right)^2} \\[5pt] % \color{black}{\left(x-\frac{1}{2}\right)^2+y^2} &=\color{black}{\left(\frac{1}{2}\right)^2} % \end{align} $$
The boundary functions of the lamina are
$$ y_{top}(x) = \color{red}{\sqrt{x(1-x)}}$$
$$ y_{bot}(x) = \color{green}{\frac{1}{2} \left(1-\sqrt{1-4 x^2}\right)}$$
The integral is $$ \int _0^{\frac{1}{2}}\int _{\frac{1}{2} \left(1-\sqrt{1-4 x^2}\right)}^{\sqrt{x-x^2}} dy dx = \int _0^{\frac{1}{2}} \sqrt{x-x^2}+\frac{1}{2} \left(\sqrt{1-4 x^2}-1\right) dx = \frac{1}{8} (\pi -2) $$
Edit: please use the more elegant solution of @Manuel Guillen.
On
Another way to look at this is via the area of a closed parametric curve, The general formula for which is given by
$$ x=f(t)\\ y=g(t)\\ A=\frac{1}{2}\int(f(t)\dot g(t)-g(t)\dot f(t))dt $$
In the present case the region is divided into two parts with different $x$ and $y$ and we have
$$A=\frac{1}{2}\left[\int_0^{\pi/4} (y_1 x_1^{'}-x_1 y_1^{'})d\theta + \int_{\pi/4}^{\pi/2} (y_2 x_2^{'}-x_2 y_2^{'})d\theta\right]$$
where
$$ x_1=\sin\theta\cos\theta;\quad y_1=\sin^2\theta\\ x_2=\cos^2\theta;\quad y_2=\sin\theta\cos\theta\\ $$
$$A=\frac{1}{2}\int_0^{\pi/4} \sin^2 \theta \, d\theta + \frac{1}{2}\int_{\pi/4}^{\pi/2} \cos^2 \theta \, d\theta= \frac{1}{8} (\pi -2)$$
as was shown by the others. The advantage here is the you are dealing with only single integrals and did not have to through all those transformations to Cartesian coordinates.
In addition, there's the trivial solution: the area under the red curve, which is $A_1=\pi(1/2)^2/4$, minus the area under the green curve, which is $A_2=(1/2)^2-A_1$. The resultant area is (again) $A=\frac{1}{8} (\pi -2)$
It is best to work in polar coordinates, as to avoid messy integrals in Cartesian coordinates with square roots of polynomials.
Looking at the graphs below:
as we go from $\theta = 0$ to $\theta = \pi/4$, we are concerned with $r = 0$ to $r = \sin \theta$. Then from $\theta = \pi/4$ to $\theta = \pi/2$, we go from $r = 0$ to $r = \cos \theta$. Also recall that: $$ dV = dx \, dy = r \, dr \, d\theta $$ So performing the area integral in polar coordinates: \begin{align*} A &= \iint dA \\ &= \int_0^{\pi/4} \int_0^{\sin \theta} r \, dr \, d\theta + \int_{\pi/4}^{\pi/2} \int_0^{\cos \theta} r \, dr \, d\theta \\ &= \frac{1}{2}\int_0^{\pi/4} \sin^2 \theta \, d\theta + \frac{1}{2}\int_{\pi/4}^{\pi/2} \cos^2 \theta \, d\theta \\ &= \frac{1}{4}\int_0^{\pi/4} (1-\cos 2\theta) \, d\theta + \frac{1}{4}\int_{\pi/4}^{\pi/2} (1+\cos 2\theta) \, d\theta \\ &= \frac{1}{4} \left[\int_0^{\pi/2} 1 \, d\theta - \int_0^{\pi/4} \cos 2\theta \, d\theta + \int_{\pi/4}^{\pi/2} \cos 2\theta \, d\theta \right] \\ &= \frac{1}{4} \left(\frac{\pi}{2} - \frac{1}{2} - \frac{1}{2} \right) \\ &= \frac{1}{4} \left(\frac{\pi}{2} - 1 \right) \\ &= \frac{\pi - 2}{8} \end{align*}