[Math] Circle integral in polar coordinates

calculuscirclesdefinite integralsintegration

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.

Best Answer

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.

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