EXAMPLE 1 Polar Coordinates Map

Describe the image of a polar rectangle \({\mathcal{R}} = [r_1,r_2]\times [\theta_1,\theta_2]\) under the polar coordinates map.

Solution Referring to Figure 15.67, we see that:

• A vertical line \(r=r_1\) (shown in red) is mapped to the set of points with radial coordinate \(r_1\) and arbitrary angle. This is the circle of radius \(r_1\).
• A horizontal line \(\theta=\theta_1\) (dashed line in the figure) is mapped to the set of points with polar angle \(\theta\) and arbitrary \(r\)-coordinate. This is the line through the origin of angle \(\theta_1\).

The image of \({\mathcal{R}} = [r_1,r_2]\times [\theta_1,\theta_2]\) under the polar coordinates map \(G(r,\theta) = (r\cos\theta, r\sin\theta)\) is the polar rectangle in the \(xy\)-plane defined by \(r_1\le r \le r_2\), \(\theta_1\le \theta \le \theta_2\).

EXAMPLE 2 Image of a Triangle

Find the image of the triangle \(\mathcal T\) with vertices \((1,2)\), \((2,1)\), \((3,4)\) under the linear map \({G}(u,v)=(2u- v,u+v)\).

Solution Because \({G}\) is linear, it maps the segment joining two vertices of \(\mathcal T\) to the segment joining the images of the two vertices. Therefore, the image of \(\mathcal T\) is the triangle whose vertices are the images Figure 15.69 \[ {G}(1,2) = (0,3),\qquad {G}(2,1) = (3,3),\qquad {G}(3,4) = (2,7) \]

Figure 15.69: The map \({G}(u,v)=(2u-v,u+v)\).

EXAMPLE 3

Let \({G}(u,v) = (uv^{-1},uv)\) for \(u>0\), \(v > 0\). Determine the images of:

• (a) The lines \(u=c\) and \(v=c\)
• (b) \([1,2]\times[1,2]\)

Find the inverse map \({G}^{-1}\).

Solution In this map, we have \(x=uv^{-1}\) and \(y=uv\). Thus \begin{equation*} xy = u^2,\qquad \frac{y}x = v^2\tag{3} \end{equation*}

• (a) By the first part of Eq. (3), \(G\) maps a point \((c,v)\) to a point in the \(xy\)-plane with \(xy=c^2\). In other words, \({G}\) maps the vertical line \(u=c\) to the hyperbola \(xy=c^2\). Similarly, by the second part of Eq. (3), the horizontal line \(v=c\) is mapped to the set of points where \(x/y=c^2\), or \(y=c^2x\), which is the line through the origin of slope \(c^2\). See Figure 15.70.
• (b) The image of \([1,2] \times [1,2]\) is the curvilinear rectangle bounded by the four curves that are the images of the lines \(u=1\), \(u=2\), and \(v=1\), \(v=2\). By Eq. (3), this region is defined by the inequalities \[ 1 \le xy \le 4,\qquad 1 \le \frac yx \le 4 \]

To find \({G}^{-1}\), we use Eq. (3) to write \(u=\sqrt{xy}\) and \(v=\sqrt{y/x}\). Therefore, the inverse map is \({G}^{-1}(x,y)=\left(\sqrt{xy}, \sqrt{y/x}\right)\). We take positive square roots because \(u>0\) and \(v > 0\) on the domain we are considering.

Figure 15.70: The mapping \({G}(u,v) = (uv^{-1},uv)\).

REMINDER

The definition of a \(2\times 2\) determinant is \begin{equation*} \left| \begin{array}{cc} a & b \\ c & d \\ \end{array} \right| = ad-bc\tag{4} \end{equation*}

EXAMPLE 4

Evaluate the Jacobian of \({G}(u,v) = (u^3+v,uv)\) at \((u,v)=(2,1)\).

Solution We have \(x=u^3+v\) and \(y=uv\), so \[ \begin{array}{rl} {\textrm{Jac}}({G})&= \frac{\partial(x,y)}{\partial(u,v)} = \left|\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array}\right|\\ &= \Big|\begin{array}{cc} 3u^2 & 1 \\ v & u \\ \end{array}\Big| = 3u^3-v \end{array} \]

The value of the Jacobian at \((2,1)\) is \({\textrm{Jac}}({G})(2,1) = 3(2)^3-1 = 23\).

THEOREM 1 Jacobian of a Linear Map

The Jacobian of a linear map \[ {G}(u,v)=(Au+Cv,Bu+Dv) \] is constant with value \begin{equation*} {\textrm{Jac}}({G}) = \left| \begin{array}{cc} A& C\\ B & D\\ \end{array} \right| = AD-BC\tag{5} \end{equation*}

Under \({G}\), the area of a region \({\mathcal{D}}\) is multiplied by the factor \(|{\textrm{Jac}}({G})|\); that is, \begin{equation*} \boxed{\textrm{Area}({G}({\mathcal{D}})) = |{\textrm{Jac}}({G})|\textrm{Area}({\mathcal{D}})}\tag{6} \end{equation*}

CONCEPTUAL INSIGHT

Although a rigorous proof of Eq. (8) is too technical to include here, we can understand Eq. (7) as an application of linear approximation. Consider a rectangle \({\mathcal{R}}\) with vertex at \(P=(u,v)\) and sides of lengths \(\Delta u\) and \(\Delta v\), assumed to be small as in Figure 15.72. The image \({G}({\mathcal{R}})\) is not a parallelogram, but it is approximated well by the parallelogram spanned by the vectors \({\bf{A}}\) and \({\bf{B}}\) in the figure: \[ \begin{array}{rl} {\bf{A}} &= {G}(u+\Delta u,v) - {G}(u,v)\\&=({x}(u+\Delta u,v)-{x}(u,v),{y}(u+\Delta u,v)-{y}(u,v)) \\ {\bf{B}} &= {G}(u,v+\Delta v) - {G}(u,v)\\ &=({x}(u,v+\Delta v)-{x}(u,v),{y}(u,v+\Delta v)-{y}(u,v)) \end{array} \]

The linear approximation applied to the components of \({G}\) yields \begin{equation*} {\bf{A}} \approx \left\langle \frac{\partial x}{\partial u}\Delta u, \frac{\partial y}{\partial u}\Delta u \right\rangle\tag{9} \end{equation*} \begin{equation*} {\bf{B}} \approx \left\langle \frac{\partial x}{\partial v}\Delta v, \frac{\partial y}{\partial v}\Delta v \right\rangle\tag{10} \end{equation*}

This yields the desired approximation: \[ \begin{array}{rl} \textrm{Area}({G}({\mathcal{R}}))&\approx \left| \det\left( \begin{array}{c} {\bf{A}}\\ {\bf{B}} \\ \end{array} \right)\right| = \left| \det\left( \begin{array}{cc} \frac{\partial x}{\partial u}\Delta u &\frac{\partial y}{\partial u}\Delta u\\ \frac{\partial x}{\partial v}\Delta v &\frac{\partial y}{\partial v}\Delta v \\ \end{array} \right)\right| \\ &= \left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}- \frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right|\, \Delta u\,\Delta v \\ & = |{\textrm{Jac}}({G})(P)|\textrm{Area}({\mathcal{R}}) \end{array} \] since the area of \({\mathcal{R}}\) is \(\Delta u\,\Delta v\).

REMINDER

Equations (9) and (10) use the linear approximations \[ \begin{array}{rl} {x}(u+\Delta u,v)-{x}(u,v)&\approx \frac{\partial x}{\partial u}\Delta u\\ {y}(u+\Delta u ,v)-{y}(u,v)&\approx \frac{\partial y}{\partial u}\Delta u \end{array} \] and \[ \begin{array}{rl} {x}(u,v+\Delta v)-{x}(u,v)&\approx \frac{\partial x}{\partial v}\Delta v\\ {y}(u,v+\Delta v)-{y}(u,v)&\approx \frac{\partial y}{\partial v}\Delta v \end{array} \]

REMINDER

\({G}\) is called “one-to-one” if \({G}(P) = {G}(Q)\) only for \(P=Q\).

THEOREM 2 Change of Variables Formula

Let \({G}:{\mathcal{D}}_0\to{\mathcal{D}}\) be a \(C^1\) mapping that is one-to-one on the interior of \({\mathcal{D}}_0\). If \(f(x,y)\) is continuous, then \begin{equation*} \boxed{\iint_{{\mathcal{D}}\,} f(x,y)\,dx\,dy = \iint_{{\mathcal{D}}_0\,} f(x(u,v),y(u,v))\left| \frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv}\tag{12} \end{equation*}

Eq. (12) is summarized by the symbolic equality \[ \boxed{dx\,dy=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv} \] Recall that \(\frac{\partial(x,y)}{\partial(u,v)}\) denotes the Jacobian \({\textrm{Jac}}({G})\).

REMINDER

If \({\mathcal{D}}\) is a domain of small diameter, \(P\in{\mathcal{D}}\) is a sample point, and \(f(x,y)\) is continuous, then (see Section 15.2) \[ \iint_{{\mathcal{D}}\,} f(x,y)\,dx\,dy \approx f(P)\textrm{Area}({\mathcal{D}}) \]

EXAMPLE 5 Polar Coordinates Revisited

Use the Change of Variables Formula to derive the formula for integration in polar coordinates.

Solution The Jacobian of the polar coordinate map \({G}(r,\theta)=(r\cos\theta, r\sin\theta)\) is \[ {\textrm{Jac}}({G})=\left| \begin{array}{cc} \displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial \theta} \\ \displaystyle\frac{\partial y}{\partial r}& \displaystyle\frac{\partial y}{\partial \theta} \\ \end{array} \right| = \left| \begin{array}{cc} \cos\theta & -r\sin\theta \\ \sin\theta& r\cos\theta \\ \end{array} \right|=r(\cos^2\theta+\sin^2\theta)=r \]

Let \({\mathcal{D}} = G({\mathcal{R}})\) be the image under the polar coordinates map \(G\) of the rectangle \({\mathcal{R}}\) defined by \(r_0\le r \le r_1\), \(\theta_0 \le \theta \le \theta_1\) (see Figure 15.75). Then Eq. (12) yields the expected formula for polar coordinates: \begin{equation*} \iint_{{\mathcal{D}}\,} f(x,y)\,dx\,dy = \int_{\theta_0}^{\theta_1}\int_{r_0}^{r_1} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\tag{13} \end{equation*}

EXAMPLE 6

Use the Change of Variables Formula to calculate \(\iint_{{\mathcal{P}}\,} e^{4x-y}\,dx\,dy\), where \({\mathcal{P}}\) is the parallelogram spanned by the vectors \(\left\langle 4,1\right\rangle\), \(\left\langle 3,3\right\rangle\) in Figure 15.75.

Figure 15.75: The map \({G}(u,v)=(4u+3v,u+3v)\).

Solution

Step 1. Define a map.

We can convert our double integral to an integral over the unit square \({\mathcal{R}}=[0,1]\times[0,1]\) if we can find a map that sends \({\mathcal{R}}\) to \({\mathcal{P}}\). The following linear map does the job: \[ \begin{array}{rl} {G}(u,v)&=(4u+3v,u+3v) \end{array} \]

Indeed, \({G}(1,0)=(4,1)\) and \({G}(0,1)=(3,3)\), so it maps \({\mathcal{R}}\) to \({\mathcal{P}}\) because linear maps map parallelograms to parallelograms.

Step 2. Compute the Jacobian.

\[ {\textrm{Jac}}({G}) = \left| \begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right| = \left| \begin{array}{cc} 4 & 3 \\ 1 & 3 \end{array} \right| = 9 \]

Step 3. Express \(f(x,y)\) in terms of the new variables.

Since \(x = 4u+3v\) and \(y=u+3v\), we have \[ e^{4x-y} = e^{4(4u+3v)-(u+3v)} = e^{15u+9v} \]

Step 4. Apply the Change of Variables Formula.

The Change of Variables Formula tells us that \(dx\,dy = 9\,du\,dv\): \[ \begin{array}{rl} \iint_{{\mathcal{P}}\,} e^{4x-y}\,dx\,dy &=\iint_{\!{\mathcal{R}}\,} e^{15u+9v}\,|{\textrm{Jac}}({G})|\,du\,dv = \int_0^1\int_0^1 e^{15u+9v}\,(9\,du\,dv)\\ &=9\left(\int_0^1\,e^{15u}\,du\right)\left(\int_0^1 e^{9v} \,dv\right) =\frac1{15}(e^{15}-1)(e^9-1) \end{array} \]

EXAMPLE 7

Use the Change of Variables Formula to compute \[ \iint_{{\mathcal{D}}\,}(x^2+y^2)\,dx\,dy \] where \({\mathcal{D}}\) is the domain \(1\le xy\le 4\), \(1\le y/x \le 4\) Figure 15.76.

Solution In Example 3, we studied the map \({G}(u,v) = (uv^{-1}, uv)\), which can be written \[ x=uv^{-1} ,\qquad y=uv \]

We showed Figure 15.76 that \({G}\) maps the rectangle \({\mathcal{D}}_0=[1,2] \times [1,2]\) to our domain \({\mathcal{D}}\). Indeed, because \(xy=u^2\) and \(xy^{-1} = v^2\), the two conditions \(1\le xy\le 4\) and \(1\le y/x \le 4\) that define \({\mathcal{D}}\) become \(1\le u\le 2\) and \(1\le v \le 2\).

The Jacobian is \[ {\textrm{Jac}}({G}) = \frac{\partial(x,y)}{\partial(u,v)} = \left| \begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{array} \right| = \left| \begin{array}{cc} v^{-1} & -uv^{-2} \\ v & u \\ \end{array} \right| = \frac{2u}v \]

To apply the Change of Variables Formula, we write \(f(x,y)\) in terms of \(u\) and \(v\): \[ f(x,y)=x^2+y^2 = \left(\frac{u}v\right)^2 + (uv)^2 = u^2(v^{-2}+v^2) \]

By the Change of Variables Formula, \[ \begin{array}{rl} \iint_{{\mathcal{D}}\,}(x^2+y^2)\,dx\,dy &= \iint_{{\mathcal{D}}_0\,}\,u^2(v^{-2}+v^2)\left|\frac{2u}v\right|\,du\,dv\\ &= 2 \int_{v=1}^2\int_{u=1}^2 u^3(v^{-3}+v)\,du\,dv\\ &= 2\left( \int_{u=1}^2u^3\,du\right)\left(\int_{v=1}^2 (v^{-3}+v) \,dv\right)\\ &= 2\left(\frac14u^4\bigg|_1^2\right)\left(-\frac12v^{-2}+\frac12v^2\bigg|_1^2\right) = \frac{225}{16} \end{array} \]

Eq. (14) can be written in the suggestive form \[ \boxed{ \frac{\partial(x,y)}{\partial(u,v)} = \left(\frac{\partial(u,v)}{\partial(x,y)}\right)^{-1} } \]

EXAMPLE 8 Using the Inverse Map

Integrate \(f(x,y) = xy(x^2+y^2)\) over \[ {\mathcal{D}}: -3\le x^2-y^2\le 3,\quad 1\le xy\le 4 \]

Solution There is a simple map \(F\) that goes in the wrong direction. Let \(u=x^2-y^2\) and \(v=xy\). Then our domain is defined by the inequalities \(-3\le u\le 3\) and \(1\le v \le 4\), and we can define a map from \({\mathcal{D}}\) to the rectangle \({\mathcal{R}} = [-3,3]\times [1,4]\) in the \(uv\)-plane Figure 15.77: \[ \begin{array}{rl} F: {\mathcal{D}} &\to {\mathcal{R}} \\ (x,y) &\to (x^2-y^2,xy) \end{array} \]

Figure 15.77: The map \(F\) goes in the “wrong” direction.

To convert the integral over \({\mathcal{D}}\) into an integral over the rectangle \({\mathcal{R}}\), we have to apply the Change of Variables Formula to the inverse mapping: \[ {G} = F^{-1}:{\mathcal{R}} \to {\mathcal{D}} \]

We will see that it is not necessary to find \({G}\) explicitly. Since \(u=x^2-y^2\) and \(v=xy\), the Jacobian of \(F\) is \[ \begin{array}{rl} {\textrm{Jac}}(F)& = \left| \begin{array}{cc} \displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y} \\ \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y} \\ \end{array} \right| = \left| \begin{array}{cc} 2x & -2y \\ y & x \\ \end{array} \right| = 2(x^2+y^2) \end{array} \]

By Eq. (14), \[ {\textrm{Jac}}({G}) = {\textrm{Jac}}(F)^{-1} = \frac1{2(x^2+y^2)} \]

Normally, the next step would be to express \(f(x,y)\) in terms of \(u\) and \(v\). We can avoid doing this in our case by observing that the Jacobian cancels with one factor of \(f(x,y)\): \[ \begin{array}{rl} \iint_{{\mathcal{D}}\,} xy(x^2+y^2) \,dx\,dy &=\iint_{{\mathcal{R}}\,} f(x(u,v), y(u,v))\,\left|{\textrm{Jac}}({G})\right|\,du\,dv \\ &= \iint_{{\mathcal{R}}\,} xy(x^2+y^2) \frac{1}{2(x^2+y^2)}\,du\,dv \\ &= \frac12 \iint_{{\mathcal{R}}\,} xy\,du\,dv \\&= \frac12 \iint_{{\mathcal{R}}\,} v\,du\,dv\qquad(\textrm{because }v=xy)\\ &= \frac12\int_{-3}^3 \int_1^4v\,dv\,du = \frac12 (6)\left(\frac124^2-\frac121^2\right) = \frac{45}2 \end{array} \]

REMINDER

\(3\times 3\)-determinants are defined in Eq. (2) of Section 12.4.