Calculus – Multiple integrals – Change of variables in multiple integrals

One-to-one transformations

\[T(u,v)=(x,y):\quad\left\{\begin{array}{l}x=g(u,v)\\[2.5mm]y=h(u,v)\end{array}\right.\quad\quad\longleftrightarrow\quad\quad T^{-1}(x,y)=(u,v):\quad\left\{\begin{array}{l}u=G(x,y)\\[2.5mm]v=H(x,y).\end{array}\right.\]

Stewart §15.9, Example 1
A transformation is defined by

\[\left\{\begin{array}{l}x=u^2-v^2\\[2.5mm]y=2uv.\end{array}\right.\]

Find the image of the square \(S=\{(u,v)\,|\,0\leq u\leq1,\;0\leq v\leq1\}\).

Solution: Start by considering the sides \(S_1\), \(S_2\), \(S_3\) and \(S_4\) of the square \(S\).

\(S_1:\;0\leq u\leq 1,\;v=0\). This implies that \(x=u^2\) and \(y=0\). So this corresponds to the line segment \(0\leq x\leq1\) and \(y=0\).

\(S_2:\;u=1,\;0\leq v\leq1\). This implies that \(x=1-v^2\) and \(y=2v\). So this corresponds to \(x=1-\frac{1}{4}y^2\) and \(0\leq x\leq1\).

\(S_3:\;0\leq u\leq1,\;v=1\). This implies that \(x=u^2-1\) and \(y=2u\). So this corresponds to \(x=\frac{1}{4}y^2-1\) and \(-1\leq x\leq0\).

\(S_4:\;u-0,\;0\leq v\leq1\). This implies that \(x=-v^2\) and \(y=0\). So this corresponds to the line segment \(-1\leq x\leq0\) and \(y=0\).

Consider the one-to-one transformation

\[T(u,v)=(x,y):\quad\left\{\begin{array}{l}x=g(u,v)\\[2.5mm]y=h(u,v)\end{array}\right.\quad\quad\longleftrightarrow\quad\quad \mathbf{r}(u,v)=g(u,v)\,\mathbf{i}+h(u,v)\,\mathbf{j}:\quad \left\{\begin{array}{l}\displaystyle\mathbf{r}_u=\frac{\partial x}{\partial u}\,\mathbf{i}+\frac{\partial y}{\partial u}\,\mathbf{j}\\[2.5mm] \displaystyle\mathbf{r}_v=\frac{\partial x}{\partial v}\,\mathbf{i}+\frac{\partial y}{\partial v}\,\mathbf{j}.\end{array}\right.\]

We can approximate the image region \(R=T(S)\) by a parallelogram determined by the secant vectors

\[\mathbf{r}(u_0+\Delta u,v_0)-\mathbf{r}(u_0,v_0)\quad\textrm{and}\quad\mathbf{r}(u_0,v_0+\Delta v)-\mathbf{r}(u_0,v_0).\]

However, we have

\[\mathbf{r}_u=\lim\limits_{\Delta u\to 0}\frac{\mathbf{r}(u_0+\Delta u,v_0)-\mathbf{r}(u_0,v_0)}{\Delta u},\]

which implies that

\[\mathbf{r}(u_0+\Delta u,v_0)-\mathbf{r}(u_0,v_0)\approx\Delta u\,\mathbf{r}_u.\]

Similarly, we have

\[\mathbf{r}(u_0,v_0+\Delta v)-\mathbf{r}(u_0,v_0)\approx\Delta v\,\mathbf{r}_v.\]

This means that we can approximate \(R\) by a parallelogram determined by the vectors \(\Delta u\,\mathbf{r}_u\) and \(\Delta v\,\mathbf{r}_v\). Therefore we can approximate the area of \(R\) by the area of this parallelogram, which is

\[\left|(\Delta u\,\mathbf{r}_u)\times(\Delta v\,\mathbf{r}_v)\right|=\left|\mathbf{r}_u\times\mathbf{r}_v\right|\,\Delta u\,\Delta v.\]

Now, if \(\mathbf{r}_u=\langle\displaystyle\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},0\rangle\) and \(\mathbf{r}_v=\langle\displaystyle\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},0\rangle\), then

\[\mathbf{r}_u\times\mathbf{r}_v=\langle0,0,\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial y}{\partial u}\\[2.5mm] \displaystyle\frac{\partial x}{\partial v} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix}\rangle =\langle0,0,\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix}\rangle.\]

So, \(|\mathbf{r}_u\times\mathbf{r}_v|\) equals the absolute value of the determinant.

Definition: The Jacobian of the transformation \(T\) given by

\[T(u,v)=(x,y):\quad\left\{\begin{array}{l}x=g(u,v)\\[2.5mm]y=h(u,v)\end{array}\right.\]

is

\[\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix}=\frac{\partial x}{\partial u}\,\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}.\]

Note: An approximation of the area \(\Delta A\) of \(R\) now is: \(\displaystyle\Delta A\approx\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,\Delta u\,\Delta v\).

Theorem: Suppose that \(T\) is a one-to-one transformation whose Jacobian is nonzero and that \(T\) maps a region \(S\) in the \(uv\)-plane onto a region \(R\) in the \(xy\)-plane. Suppose that \(f\) is continuous on \(R\) and that \(R\) and \(S\) are plane regions of type I or type II. Then:

\[\iint\limits_Rf(x,y)\,dA=\iint\limits_Sf(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv.\]

Stewart §15.9, Example 2
Use the change of variables

\[\left\{\begin{array}{l}x=u^2-v^2\\[2.5mm]y=2uv\end{array}\right.\]

to evaluate the integral \(\displaystyle\iint\limits_Ry\,dA\), where \(R\) is the region bounded by the \(x\)-axis and the parabolas \(y^2=4-4x\) and \(y^2=4+4x\) with \(y\geq0\).

Solution: We have seen that \(R=T(S)\), where \(S=[0,1]\times[0,1]\) in the \(uv\)-plane. Now we have

\[\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix} =\begin{vmatrix}2u & -2v\\[2.5mm]2v & 2u\end{vmatrix}=4(u^2+v^2).\]

Hence we have

\begin{align*} \iint\limits_Ry\,dA&=\iint\limits_S2uv\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv =\int_0^1\int_0^12uv\cdot4(u^2+v^2)\,du\,dv\\[2.5mm] &=8\int_0^1\int_0^1(u^3v+uv^3)\,du\,dv=8\int_0^1\bigg[\frac{1}{4}u^4v+\frac{1}{2}u^2v^3\bigg]_{u=0}^1\,dv\\[2.5mm] &=\int_0^1\left(2v+4v^3\right)\,dv=\bigg[v^2+v^4\bigg]_{v=0}^1=2. \end{align*}

Stewart §15.9, Example 3
Evaluate the integral \(\displaystyle\iint\limits_Re^{(x+y)/(x-y)}\,dA\), where \(R\) is the trapezoidal region with vertices \((1,0)\), \((2,0)\), \((0,-2)\) and \((0,-1)\).

Solution: Consider the change of variables \(u=x+y\) and \(v=x-y\), which implies that \(x=\frac{1}{2}(u+v)\) and \(y=\frac{1}{2}(u-v)\). Then the Jacobian is

\[\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix} =\begin{vmatrix}\frac{1}{2} & \frac{1}{2}\\[2.5mm]\frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}.\]

Note that

\[\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}\displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y}\\[2.5mm] \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y}\end{vmatrix} =\begin{vmatrix}1 & 1\\[2.5mm]1 & -1\end{vmatrix}=-2\]

which implies that

\[\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\displaystyle\frac{\partial(u,v)}{\partial(x,y)}}=-\frac{1}{2}.\]

To find the region \(S\) in the \(uv\)-plane corresponding to \(R\), we note that the sides of \(R\) lie on the lines \(y=0\), \(x-y=2\), \(x=0\) and \(x-y=1\). These lines correspond to \(u=v\), \(v=2\), \(u=-v\) and \(v=1\) respectively. This implies that \(S\) is the trapezoidal region with vertices \((1,1)\), \((2,2)\), \((-2,2)\) and \((-1,1)\):

\[S=\{(u,v)\,|\,1\leq v\leq2,\;-v\leq u\leq v\}.\]

Hence we have

\begin{align*} \iint\limits_Re^{(x+y)/(x-y)}\,dA&=\iint\limits_Se^{u/v}\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv =\int_1^2\int_{-v}^ve^{u/v}\cdot\frac{1}{2}\,du\,dv=\frac{1}{2}\int_1^2\bigg[ve^{u/v}\bigg]_{u=-v}^v\,dv\\[2.5mm] &=\frac{1}{2}\int_1^2\left(e-e^{-1}\right)v\,dv =\frac{1}{2}\left(e-e^{-1}\right)\bigg[\frac{1}{2}v^2\bigg]_{v=1}^2=\frac{3}{4}\left(e-e^{-1}\right). \end{align*}

Polar coordinates

In the case of polar coordinates we have: \(\displaystyle T(r,\theta)=(x,y):\quad\left\{\begin{array}{l}x=r\cos(\theta)\\[2.5mm] y=r\sin(\theta).\end{array}\right.\)

The Jacobian of \(T\) is

\[\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial\theta}\\[2.5mm] \displaystyle\frac{\partial y}{\partial r} & \displaystyle\frac{\partial y}{\partial\theta}\end{vmatrix} =\left|\begin{array}{cc}\cos(\theta) & -\sin(\theta)\\[2.5mm]r\sin(\theta) & r\cos(\theta)\end{array}\right|=r>0.\]

Note that the rectangle

\[S=\{(r,\theta)\,|\,a\leq r\leq b,\;\alpha\leq\theta\leq\beta\}\]

in the \(r\theta\)-plane corresponds to the polar rectangle

\[R=\{(r\cos(\theta),r\sin(\theta))\,|\,a\leq r\leq b,\;\alpha\leq\theta\leq\beta\}\]

in the \(xy\)-plane. Hence we have

\begin{align*} \iint\limits_Rf(x,y)\,dA&=\iint\limits_Sf(r\cos(\theta),r\sin(\theta))\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|\,dr\,d\theta\\[2.5mm] &=\int_{\alpha}^{\beta}\int_a^bf(r\cos(\theta),r\sin(\theta))\,r\,dr\,d\theta. \end{align*}

Triple integrals

The Jacobian of the transformation \(T\) given by

\[T(u,v,w)=(x,y,z):\quad\left\{\begin{array}{l}x=g(u,v,w)\\[2.5mm]y=h(u,v,w)\\[2.5mm]z=k(u,v,w)\end{array}\right.\]

is

\[\frac{\partial(x,y,z)}{\partial(u,v,w)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v} & \displaystyle\frac{\partial x}{\partial w}\\[2.5mm] \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v} & \displaystyle\frac{\partial y}{\partial w}\\[2.5mm] \displaystyle\frac{\partial z}{\partial u} & \displaystyle\frac{\partial z}{\partial v} & \displaystyle\frac{\partial z}{\partial w}\end{vmatrix}.\]

Such a determinant is defined as a scalar triple product (see also the course on Linear Algebra for more details):

\[\begin{vmatrix}a_1&b_1&c_1\\[2.5mm]a_2&b_2&c_2\\[2.5mm]a_3&b_3&c_3\end{vmatrix} =\begin{vmatrix}a_1&a_2&a_3\\[2.5mm]b_1&b_2&b_3\\[2.5mm]c_1&c_2&c_3\end{vmatrix} =\det(\mathbf{a},\mathbf{b},\mathbf{c})=\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right).\]

Then we have

\[\iiint\limits_Rf(x,y,z)\,dV=\iiint\limits_Sf(x(u,v,w),y(u,v,w),z(u,v,w))\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|\,du\,dv\,dw.\]

Cylindrical coordinates

In the case of cylindrical coordinates we have: \(\displaystyle T(r,\theta,z)=(x,y,z):\quad \left\{\begin{array}{l}x=r\cos(\theta)\\[2.5mm]y=r\sin(\theta)\\[2.5mm]z=z.\end{array}\right.\)

The Jacobian of \(T\) is (see the course on Linear Algebra for details about the computation)

\[\frac{\partial(x,y,z)}{\partial(r,\theta,z)}=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial\theta} & \displaystyle\frac{\partial x}{\partial z}\\[2.5mm] \displaystyle\frac{\partial y}{\partial r} & \displaystyle\frac{\partial y}{\partial\theta} & \displaystyle\frac{\partial y}{\partial z}\\[2.5mm] \displaystyle\frac{\partial z}{\partial r} & \displaystyle\frac{\partial z}{\partial\theta} & \displaystyle\frac{\partial z}{\partial z}\end{vmatrix} =\begin{vmatrix}\cos(\theta) & -\sin(\theta) & 0\\[2.5mm]r\sin(\theta) & r\cos(\theta) & 0\\[2.5mm]0 & 0 & 1\end{vmatrix} =\begin{vmatrix}\cos(\theta) & -\sin(\theta)\\[2.5mm]r\sin(\theta) & r\cos(\theta)\end{vmatrix}=r>0.\]

Spherical coordinates

In the case of spherical coordinates we have: \(\displaystyle T(\rho,\phi,\theta)=(x,y,z):\quad \left\{\begin{array}{l}x=\rho\sin(\phi)\cos(\theta)\\[2.5mm]y=\rho\sin(\phi)\sin(\theta)\\[2.5mm]z=\rho\cos(\phi).\end{array}\right.\)

The Jacobian of \(T\) is (see the course on Linear Algebra for details about the computation)

\begin{align*} \frac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)}&=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial\rho} & \displaystyle\frac{\partial x}{\partial\phi} & \displaystyle\frac{\partial x}{\partial\theta}\\[2.5mm] \displaystyle\frac{\partial y}{\partial\rho} & \displaystyle\frac{\partial y}{\partial\phi} & \displaystyle\frac{\partial y}{\partial\theta}\\[2.5mm] \displaystyle\frac{\partial z}{\partial\rho} & \displaystyle\frac{\partial z}{\partial\phi} & \displaystyle\frac{\partial z}{\partial\theta}\end{vmatrix} =\begin{vmatrix}\sin(\phi)\cos(\theta) & \rho\cos(\phi)\cos(\theta) & -\rho\sin(\phi)\sin(\theta)\\[2.5mm] \sin(\phi)\sin(\theta) & \rho\cos(\phi)\sin(\theta) & \rho\sin(\phi)\cos(\theta)\\[2.5mm]\cos(\phi) & -\rho\sin(\phi) & 0\end{vmatrix}\\[2.5mm] &=\cos(\phi)\begin{vmatrix}\rho\cos(\phi)\cos(\theta) & -\rho\sin(\phi)\sin(\theta)\\[2.5mm] \rho\cos(\phi)\sin(\theta) & \rho\sin(\phi)\cos(\theta)\end{vmatrix}+\rho\sin(\phi)\begin{vmatrix}\sin(\phi)\cos(\theta) & -\rho\sin(\phi)\sin(\theta)\\[2.5mm] \sin(\phi)\sin(\theta) & \rho\sin(\phi)\cos(\theta)\end{vmatrix}\\[2.5mm] &=\cos(\phi)\left(\rho^2\sin(\phi)\cos(\phi)\cos^2(\theta)+\rho^2\sin(\phi)\cos(\phi)\sin^2(\theta)\right) +\rho\sin(\phi)\left(\rho\sin^2(\phi)\cos^2(\theta)+\rho\sin^2(\phi)\sin^2(\theta)\right)\\[2.5mm] &=\cos(\phi)\cdot\rho^2\sin(\phi)\cos(\phi)+\rho\sin(\phi)\cdot\rho\sin^2(\phi) =\rho^2\sin(\phi)\left(\cos^2(\phi)+\sin^2(\phi)\right)=\rho^2\sin(\phi)>0. \end{align*}

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Last modified on October 5, 2021