Change of variables in double integrals

Multiple integrals: Double integrals

Change of variables in double integrals

Suppose $x=x(u,v)$ and $y=y(u,v)$ is an invertible map of a region $S$ in the $uv$ -plane to a region $R$ in the $xy$ -plane and suppose the functions $x$ and $y$ have continuous derivatives on $S$ . If the double integral $\displaystyle \iint_R f(x,y)\,\dd(x,y)$ exists (i.e. $f(x,y)$ is integrable on $R$ ) and if $g(u,v)=f\bigl(x(u,v), y(u,v)\bigr)$ , then $g(u,v)$ is integrable on $S$ and

$\iint_R f(x,y)\,\dd(x,y) =\iint_S g(u,v)\cdot \bigl|J(u,v)\bigr|\,\dd u\, \dd v)$ where the Jacobian $J(u,v)$ is defined as

$\begin{aligned}J(u,v)&=\det\matrix{\dfrac{\partial x}{\partial u} &\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}}\\[0.25cm] &= \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} -\frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}\end{aligned}$

With polar coordinates we have $x(r,\varphi)=r\cos\varphi$ and $y(r,\varphi)=r\sin\varphi$ . Then:

$\begin{aligned}J(r,\varphi) &= \det\matrix{\dfrac{\partial x}{\partial r} &\dfrac{\partial x}{\partial \varphi}\\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \varphi}}\\[0.25cm] &=\det\matrix{\cos\varphi & -r\sin\varphi\\ \sin\varphi & r\cos\varphi} \\[0.25cm] &= r(\cos^2\varphi+\sin^2\varphi)\\[0.25cm] &= r\end{aligned}$

With the coordinate transformation $(u,v)\mapsto \bigl(x(u,v), y(u,v)\bigr)$ a rectangular region in the $uv$ -plane, let's call it $S$ , is mapped onto a not necessarily rectangular region in the $xy$ -plane, let's call it $R$ . To compute the double integral of a function over $S$ we overlay $S$ with a rectangular partition with mesh sizes $\dd u$ and $\dd v$ . Choose a fixed vertex $(u_0,v_0)$ in a sub-rectangle of the partition. The vertices $P=(u_0,v_0)$ , $Q=(u_0+\dd u, v_0)$ and $R=(u_0, v_0+\dd v)$ are mapped to $\bigl(x_0,y_0\bigr)=\bigl(x(u_0, v_0), y(u_0, v_0)\bigr)$ $\bigl(x(u_0+\dd u, v_0), y(u_0+\dd u, v_0)\bigr)$ and $\bigl(x(u_0,v_0+\dd u), y(v_0,u_0+\dd u)\bigr)$ , respectively. The transformation is visualised below.

dubbelintegraal in andere coördinaten

The sub-rectangle in the $uv$ -plane is approximately mapped to a two-sided rhombus given by the vectors $\vec{PQ}$ and $\vec{PR}$ . In the figure below we show that it is an approximation, but in the infinitesimal case it is correct.

transformatie van deelgebiedje

In the infinitesimal case we have:

$\vec{PQ}=\cv{\dfrac{\partial x}{\partial u} \\ \dfrac{\partial y}{\partial u}}\cdot \dd u\qquad\text{and} \qquad \vec{PR}=\cv{\dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial v}}\cdot \dd v$ The area of the rhombus is the length of the cross product of the vectors $\vec{PQ}$ and $\vec{PR}$ . This is equal to $\bigl|J(u,v)\bigr|\,\dd u\,\dd v$ .

We now know how a rectangular sub-region is transformed. Hereafter we still have to do summation over the partition and take a limit to complete the proof.

Calculate the double integral

$\iint_R e^{\frac{x-y}{x+y}}\,\dd(x,y)$ where the region of integration $R$ is the triangle enclosed by the coordinate axes and the line $y+x=1$ .

$\displaystyle\iint_R e^{\frac{x-y}{x+y}}\,\dd(x,y)={}$ $\frac{1}{4}\left(e-\frac{1}{e}\right)$

By the substitution

$u=x+y\quad\text{and}\quad v=x-y$ the integrand becomes a simpler expression, which is $e^{\frac{v}{u}}$ . The region of integration is mapped to a new triangle $S$ in the $uv$ -plane bounded by the lines $u=v$ , $u=-v$ and $u=1$ . After all, we can isolate $x$ and $y$ in the above equations as

$x=\frac{1}{2}(u+v)\quad\text{and}\quad y=\frac{1}{2}(u-v)$ Therefore:

$\begin{aligned} y=-x+1&\iff u=1\\[0.25cm] x=0 &\iff u=-v\\[0.25cm] y=0 &\iff u=v\end{aligned}$ With this we then have the coordinate transformation that maps $S$ to $R$ in the $xy$ -plane. The Jacobian $J(u,v)$ can also be calculated directly:

$\begin{aligned}J(u,v) &= \det\matrix{\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2}}\\[0.25cm] &= -\frac{1}{2}\end{aligned}$ The calculation of the double integral now proceeds as follows:

$\begin{aligned} \iint_R e^{\frac{x-y}{x+y}}\,\dd(x,y) &= \iint_S e^{\frac{v}{u}}\,\bigl|J(u,v)\bigr|\,\dd(u,v) \\[0.25cm] &= \frac{1}{2}\int_{u=0}^{u=1}\left(\int_{v=-u}^{v=u}e^{\frac{v}{u}}\,\dd v\right)\dd u\\[0.25cm] &= \frac{1}{2}\int_{u=0}^{u=1} \biggl[u\,e^{\frac{v}{u}}\biggr]_{v=-u}^{v=u}\;\dd u\\[0.25cm] &=\frac{1}{2}\int_{u=0}^{u=1}\left(e-\frac{1}{e}\right)u\,\dd u\\[0.25cm] &= \frac{1}{2}\left(e-\frac{1}{e}\right)\cdot\biggl[\frac{1}{2}u^2\biggr]_{u=0}^{u=1}\\[0.25cm] &= \frac{1}{4}\left(e-\frac{1}{e}\right)\end{aligned}$

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