# How to Draw the R region of integration and change the order of integration (do NOT evaluate the integral)?

## ${\int}_{0}^{1} {\int}_{-} {\sqrt{1 - {y}^{2}}}^{\sqrt{1 - {y}^{2}}} \mathrm{dx} \mathrm{dy}$

Jul 30, 2018

This integration is all about the unit circle, because the underlying relationship is:

• ${x}^{2} + {y}^{2} = 1 \implies \left\{\begin{matrix}x = \pm \sqrt{1 - {y}^{2}} \\ y = \pm \sqrt{1 - {x}^{2}}\end{matrix}\right.$

Here, because $R$ is defined as follows, it is specifically about the area within Q1 and Q2:

• $\left\{\begin{matrix}- \sqrt{1 - {y}^{2}} \le x \le \sqrt{1 - {y}^{2}} \\ q \quad q \quad q \quad 0 \le y \le 1\end{matrix}\right.$

The order can be reversed as:

• $\left\{\begin{matrix}0 \le y \le \sqrt{1 - {x}^{2}} \\ - 1 \le x \le 1\end{matrix}\right.$

So:

• ${\int}_{0}^{1} {\int}_{-} {\sqrt{1 - {y}^{2}}}^{\sqrt{1 - {y}^{2}}} \setminus \mathrm{dx} \setminus \mathrm{dy} = {\int}_{-} {1}^{1} {\int}_{0}^{\sqrt{1 - {x}^{2}}} \setminus \mathrm{dy} \setminus \mathrm{dx}$