# How to change the limits of a double integral to polar coordinates limits?

Problem:

Use polar coordinates to evaluate the following integral:

$$\int_{0}^{2}\int_{0}^{\sqrt{2x-x^{2}}}xdydy$$

Solution:

First, this is the graph I manually plotted to define the new limits:

So I set up the new integral with these new limits in polar coordinates:

$$1\leqslant r\leqslant 2$$

$$0\leqslant \theta\leqslant\pi$$

But the integral gives me $0$ as a result. This is the integral to evaluate:

$$\int_{0}^{\pi}\int_{1}^{2}r\cos(\theta)rdrd\theta$$

## 5 Answers

The problem is your range for $r$. In this case, the circle has equation $$(x-1)^2+y^2=1,$$from where you take $$x^2-2x+1+y^2=1 \Leftrightarrow (x^2+y^2)=2x \Leftrightarrow r^2=2r\cos(\theta) \Leftrightarrow r=2\cos(\theta)$$ so actually $r\in[0,2\cos(\theta)]$.

The problem is that polar coordinates are centered at the origin, and your original region of integration is not.

To fix this, I'd start by making the change of variables $w = x-1$ in the original integral, which results in $$\int_{-1}^1 \int_{0}^\sqrt{1-w^2} (w + 1) \; dy \; dw.$$

Now convert to polar to get $$\int_0^\pi \int_0^1 (r \cos(\theta) + 1) r \; dr \; d\theta$$

Notice that

$$2x - x^2 = (1 - 1) + 2x - x^2 = 1 - (x-1)^2 ,$$

so $0 \le y \le \sqrt {2x - x^2}$ implies $y^2 \le 1 - (x-1)^2$ (this is the equation of a circle!), which in turn implies $y^2 + (x-1)^2 \le 1$, so you have to choose polar coordinates relative to the circle of radius $1$ and center $(1,0)$. This means

$$x = \color{red} {1 + } r \cos \theta \\ y = r \sin \theta$$

with $0 \le r \le 1$ and $0 \le \theta \le \pi$. With this, you should be able to find your way by yourself now.

Your conversion from cartesian bounds to polar is not quite correct. Your bounds describe an area that looks like this:

One way to go about this is to substitute $x$ with a variable $u = x - 1$ and then integrate over the bounds $\int_0^\pi\int_0^1$ after converting $\int_0^2\int_0^{\sqrt{2(u+1) - (u+1)^2}}(u+1)u\mathrm{d}u\mathrm{d}y = \int_0^2\int_0^{\sqrt{1 - u^2}}(u+1)\mathrm{d}u\mathrm{d}y$

The region in your sketch is half a tangent circle. The equation is not $r=2$, but $r=2\cos \theta$. Integrate $\theta$ from $0$ to $\pi/2$ and $r$ from 0 to $2\cos \theta.$