LEC CAL2 8

Change of variables

If $g:[a,b]\to \mathbb{R}$ is continuously differentiable and $f:\mathbb{R}\to \mathbb{R}$ is continuous, then

$${\int }_{g(a)}^{g(b)}f={\int }_a^b(f\circ g)\cdot g^{\prime }.$$

[!Theorem]
Let $A\subseteq {\mathbb{R}}^n$ be open and let $g:A\to {\mathbb{R}}^n$ be a 1-1, continuously differentiable function such that $\det g^{\prime }(x)\ne 0$ for all $x\in A$. Suppose that $f:g(A)\to \mathbb{R}$ is integrable¹. Then,

$${\int }_{g(A)}f={\int }_A(f\circ g)|\det g^{\prime }|.$$

[!Proof]-

[!Example]
Evaluate:

$${\int }_0^1{\int }_0^{\sqrt{1-x^2}}{e}^{1-x^2-y^2}dydx.$$

$g$ will be of type $(0,1)\times (0,\pi /2)\to {\mathbb{R}}^2$, $g(r,\theta )=(r\cos \theta ,r\sin \theta )$.

$$g^{\prime }(r,\theta )=\left[\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right]$$

$\det g^{\prime }=r$. Thus, the above integral is equal to

$${\int }_A{e}^{1-r^2}r={\int }_0^{\pi /2}{\int }_0^{1}r{e}^{1-r^2}drd\theta =\pi (e-1)/4$$

[!Example]
Find the area of the region in ${\mathbb{R}}^2$ bounded by $y=x^3$, $y=2x^3$, $xy=1$, $xy=3$.

Use change of variables $u=x^3/y$, $v=xy$. $g(u,v)=(u^{1/4}{v}^{1/4},u^{-1/4}{v}^{3/4})$. The region $A$ is $(1/2,1)\times (1,3)$ (we are only looking at the first quadrant).

$$g^{\prime }(u,v)=1/4\left[\begin{array}{cc}{u}^{-3/4}{v}^{1/4}& u^{1/4}{v}^{-3/4}\\ -u^{-5/4}{v}^{3/4}& 3u^{-1/4}{v}^{-1/4}\end{array}\right]$$

$\det g^{\prime }=\frac{1}{4u}$. Thus, the area is given by

$$\frac{1}{4}{\int }_1^3{\int }_{1/2}^1\frac{1}{u}dudv=\frac{\ln 2}{2}.$$

The total area is $\ln 2$.

Footnotes

1. Note that $g(A)$ is open. ↩