Circle integral in polar coordinates I know how to integrate and deduce the area of a circle using vertical "slices" (dx). However, I wanted to know how to solve the problem another way: slicing the circle like one would slice a cake, and summing the small triangular areas. I reached the following expressions: 
$$
\text{Area of a circle} = 2\int_0^π r^2\sin(θ)\cos(θ) \,dθ
 = 2r^2\int_0^π \sin(θ)\cos(θ) \,dθ.
$$
where $r\sin(θ)$ and $r\cos(θ)$ are the sides of the little triangles which I'm summing.
Are the formulas I reached correct? And if so, how do I solve the integral?
Thank you.
 A: The area of the triangle within $(r,\theta)$ and $(r,\theta+d\theta)$ is $r\,d\theta$. The area of the circle (constant $r$) in polar coordinates is
$$\int_0^{2\pi}\int_0^{R}r\,drd\theta=\frac{R^2}{2}\int_0^{2\pi}d\theta=\pi R^2$$
The area of the triangle is not how you represent it, you've given the points on the circle, i.e. $(r,\theta)$ and $(r,\theta+d\theta)$, so you are not integrating a differential area.
A: Your idea is good, but some details are shady. If a curve's radius function can be expressed as a function of its angle with the positive side of the x-axis, $r(\theta)$, then the area of the curve between two half-lines $\theta=\alpha$ and $\theta = \beta$ is $A=\frac{1}{2}\int_{\alpha}^{\beta}r^2(\theta)d\theta.$ This can be proven formally, but it can be seen intuitively as the sum of many infinitesimally small arcs of radius $r(\theta)$, and angle $d\theta$, so the areas of those arcs are approximately (using the area of the triangle) $\frac{1}{2}(r(\theta))^2\sin{d\theta}$. Since $d\theta$ is infinitesimally small, we have $\sin{d\theta} \approx d\theta$, so the area of the small triangle is approximately $\frac{1}{2}r^2d\theta$.
Therefore, the area of a circle of radius $r$ is ($r(\theta)=r$) $A=\frac{1}{2}\int_{0}^{2\pi}r^2d\theta=r^2\pi.$
