5.7 Change of variables in multiple integrals

Recall from Substitution Rule the method of integration by substitution. When evaluating an integral such as ${\displaystyle {\int }_2^{3}x{(x^2-4)}^{5}dx,}$ we substitute $u=g\left(x\right)=x^2-4.$ Then $du=2xdx$ or $xdx=\frac{1}{2}du$ and the limits change to $u=g(2)=2^2-4=0$ and $u=g\left(3\right)=9-4=5.$ Thus the integral becomes ${\displaystyle {\int }_0^5\frac{1}{2}{u}^{5}du}$ and this integral is much simpler to evaluate. In other words, when solving integration problems, we make appropriate substitutions to obtain an integral that becomes much simpler than the original integral.

We also used this idea when we transformed double integrals in rectangular coordinates to polar coordinates and transformed triple integrals in rectangular coordinates to cylindrical or spherical coordinates to make the computations simpler. More generally,

Where $x=g\left(u\right),dx=g\prime \left(u\right)du,$ and $u=c$ and $u=d$ satisfy $c=g\left(a\right)$ and $d=g\left(b\right).$

A similar result occurs in double integrals when we substitute $x=f\left(r,\theta \right)=r\text{cos}\theta ,$ $y=g\left(r,\theta \right)=r\text{sin}\theta ,$ and $dA=dxdy=rdrd\theta .$ Then we get

where the domain $R$ is replaced by the domain $S$ in polar coordinates. Generally, the function that we use to change the variables to make the integration simpler is called a transformation    or mapping.

Planar transformations

A planar transformation     $T$ is a function that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables. Both $G$ and $R$ are subsets of $R^2.$ For example, [link] shows a region $G$ in the $uv\text{-plane}$ transformed into a region $R$ in the $xy\text{-plane}$ by the change of variables $x=g\left(u,v\right)$ and $y=h\left(u,v\right),$ or sometimes we write $x=x\left(u,v\right)$ and $y=y\left(u,v\right).$ We shall typically assume that each of these functions has continuous first partial derivatives, which means $g_u,g_v,h_u,$ and $h_v$ exist and are also continuous. The need for this requirement will become clear soon.

To show that $T$ is a one-to-one transformation, we assume $T\left(u_1,v_1\right)=T\left(u_2,v_2\right)$ and show that as a consequence we obtain $\left(u_1,v_1\right)=\left(u_2,v_2\right).$ If the transformation $T$ is one-to-one in the domain $G,$ then the inverse $T^{\mathrm{-1}}$ exists with the domain $R$ such that $T^{\mathrm{-1}}\circ T$ and $T\circ T^{\mathrm{-1}}$ are identity functions.

[link] shows the mapping $T\left(u,v\right)=\left(x,y\right)$ where $x$ and $y$ are related to $u$ and $v$ by the equations $x=g\left(u,v\right)$ and $y=h\left(u,v\right).$ The region $G$ is the domain of $T$ and the region $R$ is the range of $T,$ also known as the image of $G$ under the transformation $T.$

Determining how the transformation works

Suppose a transformation $T$ is defined as $T\left(r,\theta \right)=\left(x,y\right)$ where $x=r\text{cos}\theta ,y=r\text{sin}\theta .$ Find the image of the polar rectangle $G=\left\{\left(r,\theta \right)|0<r\le 1,0\le \theta \le \pi \text{/}2\right\}$ in the $r\theta \text{-plane}$ to a region $R$ in the $xy\text{-plane}\text{.}$ Show that $T$ is a one-to-one transformation in $G$ and find $T^{\mathrm{-1}}\left(x,y\right).$

Since $r$ varies from 0 to 1 in the $r\theta \text{-plane},$ we have a circular disc of radius 0 to 1 in the $xy\text{-plane}\text{.}$ Because $\theta$ varies from 0 to $\pi \text{/2}$ in the $r\theta \text{-plane},$ we end up getting a quarter circle of radius $1$ in the first quadrant of the $xy\text{-plane}$ ( [link] ). Hence $R$ is a quarter circle bounded by $x^2+y^2=1$ in the first quadrant.

In order to show that $T$ is a one-to-one transformation, assume $T\left(r_1,{\theta }_1\right)=T\left(r_2,{\theta }_2\right)$ and show as a consequence that $\left(r_1,{\theta }_1\right)=\left(r_2,{\theta }_2\right).$ In this case, we have

$\begin{array}{ccc}\hfill T\left(r_1,{\theta }_1\right)& =\hfill & T\left(r_2,{\theta }_2\right),\hfill \\ \hfill \left(x_1,y_1\right)& =\hfill & \left(x_1,y_1\right),\hfill \\ \hfill \left(r_1\text{cos}{\theta }_1,r_1\text{sin}{\theta }_1\right)& =\hfill & \left(r_2\text{cos}{\theta }_2,r_2\text{sin}{\theta }_2\right),\hfill \\ \hfill r_1\text{cos}{\theta }_1& =\hfill & r_2\text{cos}{\theta }_2,r_1\text{sin}{\theta }_1=r_2\text{sin}{\theta }_2.\hfill \end{array}$

Dividing, we obtain

$\begin{array}{ccc}\hfill \frac{r_1\text{cos}{\theta }_1}{r_1\text{sin}{\theta }_1}& =\hfill & \frac{r_2\text{cos}{\theta }_2}{r_2\text{sin}{\theta }_2}\hfill \\ \hfill \frac{\text{cos}{\theta }_1}{\text{sin}{\theta }_1}& =\hfill & \frac{\text{cos}{\theta }_2}{\text{sin}{\theta }_2}\hfill \\ \hfill \text{tan}{\theta }_1& =\hfill & \text{tan}{\theta }_2\hfill \\ \hfill {\theta }_1& =\hfill & {\theta }_2\hfill \end{array}$

since the tangent function is one-one function in the interval $0\le \theta \le \pi \text{/}2.$ Also, since $0<r\le 1,$ we have $r_1=r_2,{\theta }_1={\theta }_2.$ Therefore, $\left(r_1,{\theta }_1\right)=\left(r_2,{\theta }_2\right)$ and $T$ is a one-to-one transformation from $G$ into $R.$

To find $T^{\mathrm{-1}}\left(x,y\right)$ solve for $r,\theta$ in terms of $x,y.$ We already know that $r^2=x^2+y^2$ and $\text{tan}\theta =\frac{y}{x}.$ Thus $T^{\mathrm{-1}}\left(x,y\right)=\left(r,\theta \right)$ is defined as $r=\sqrt{x^2+y^2}$ and $\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{y}{x}\right).$

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