Math24.net Calculus
 Home Calculus Limits and Continuity Differentiation Applications of Derivative Integration Sequences and Series Double Integrals Triple Integrals Line Integrals Surface Integrals Fourier Series Differential Equations 1st Order Equations 2nd Order Equations Nth Order Equations Systems of Equations Formulas and Tables
Change of Variables in Double Integrals
Sometimes, it is often advantageous to evaluate $$\iint\limits_R {f\left( {x,y} \right)dxdy}$$ in a coordinate system other than the $$xy$$-coordinate system. This may be as a consequence either of the shape of the region, or of the complexity of the integrand. Calculating the double integral in the new coordinate system can be much simpler.

The formula for change of variables is given by ${\iint\limits_R {f\left( {x,y} \right)dxdy} } = {\iint\limits_S {f\left[ {x\left( {u,v} \right),y\left( {u,v} \right)} \right]\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dxdy} ,}$ where the expression $$\left| {\large\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}\normalsize} \right| = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\ {\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}} \end{array}} \right| \ne 0$$ is the so-called Jacobian of the transformation $$\left( {x,y} \right) \to \left( {u,v} \right),$$ and $$S$$ is the pullback of the region of integration $$R$$ which can be computed by substituting $$x = x\left( {u,v} \right),$$ $$y = y\left( {u,v} \right)$$ into the definition of $$R.$$ Notice, that $$\left| {\large\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}\normalsize} \right|$$ in the formula above means the absolute value of the corresponding determinant.

Supposing that the transformation $$\left( {x,y} \right) \to \left( {u,v} \right)$$ is a $$1-1$$ mapping from $$R$$ to a region $$S,$$ the inverse relation is described by the Jacobian $\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {{{\left( {\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}} \right)}^{ - 1}}} \right|.$
Thus, use of change of variables in a double integral requires the following 3 steps:
1. Find the pulback $$S$$ in the new coordinate system $$\left( {u,v} \right)$$ for the initial region of integration $$R;$$

2. Calculate the Jacobian of the transformation $$\left( {x,y} \right) \to \left( {u,v} \right)$$ and write down the differential through the new variables: $$dxdy = \left| {\large\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}\normalsize} \right|dudv;$$

3. Replace $$x$$ and $$y$$ in the integrand by substituting $$x = x\left( {u,v} \right)$$ and $$y = y\left( {u,v} \right),$$ respectively.

Example 1
Calculate the double integral $$\iint\limits_R {\left( {y - x} \right)dxdy},$$ where the region $$R$$ is bounded by $$y = x + 1,$$ $$y = x - 3,$$ $$y = - {\large\frac{x}{3}\normalsize} + 2,$$ $$y = - {\large\frac{x}{3}\normalsize} + 4.$$

Solution.
The sketch of the region $$R$$ is given in Figure $$1.$$ We use change of variables to simplify the integral. By letting $$u = y - x,$$ $$v = y + {\large\frac{x}{3}\normalsize},$$ we have ${y = x + 1,}\;\; {\Rightarrow y - x = 1,}\;\; {\Rightarrow u = 1,}$ ${y = x - 3,}\;\; {\Rightarrow y - x = -3,}\;\; {\Rightarrow u = -3,}$ ${y = - \frac{x}{3} + 2,}\;\; {\Rightarrow y + \frac{x}{3} = 2,}\;\; {\Rightarrow v = 2,}$ ${y = - \frac{x}{3} + 4,}\;\; {\Rightarrow y + \frac{x}{3} = 4,}\;\; {\Rightarrow v = 4.}$ Hence, the pullback $$S$$ of the region $$R$$ is the rectangle shown in Figure $$2.$$
 Fig.1 Fig.2
Calculate the Jacobian of this transformation. ${\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}\\ {\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial \left( {y - x} \right)}}{{\partial x}}}&{\frac{{\partial \left( {y - x} \right)}}{{\partial y}}}\\ {\frac{{\partial \left( {y + \frac{x}{3}} \right)}}{{\partial x}}}&{\frac{{\partial \left( {y + \frac{x}{3}} \right)}}{{\partial y}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} { - 1}&1\\ {\frac{1}{3}}&1 \end{array}} \right| } = { - 1 \cdot 1 - 1 \cdot \frac{1}{3} = - \frac{4}{3}.}$ Then the absolute value of the Jacobian is ${\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| } = {\left| {{{\left( {\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}} \right)}^{ - 1}}} \right| } = {\left| {\frac{1}{{ - \frac{4}{3}}}} \right| = \frac{3}{4}.}$ Hence, the differential is ${dxdy = \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv } = {\frac{3}{4}dudv.}$ so calculating the integral in the new variables $$\left( {u,v} \right)$$ is much simpler: ${\iint\limits_R {\left( {y - x} \right)dxdy} } = {\iint\limits_S {\left( u \cdot \frac{3}{4}dudv \right)} } = {\frac{3}{4}\int\limits_{ - 3}^1 {udu} \int\limits_2^4 {dv} } = {\frac{3}{4}\left. {\left( {\frac{{{u^2}}}{2}} \right)} \right|_{ - 3}^1 \cdot \left. v \right|_2^4 } = {\frac{3}{4}\left( {\frac{1}{2} - \frac{9}{2}} \right) \cdot \left( {4 - 2} \right) = - 6.}$
Example 2
Evaluate the double integral $$\iint\limits_R {\left( {x + y} \right)dxdy},$$ where the region of integration $$R$$ is bounded by the lines $$y = x,$$ $$y = 2x,$$ $$x + y = 2.$$

Solution.
The region $$R$$ is an irregular triangle and shown in Figure $$3.$$ To simplify the region of integration, we make the following substitution: $$y - x = u,$$ $$y - 2x = v.$$ Next, we express $$x, y$$ as functions of $$u, v$$ and define the pullback $$S$$ of the region of integration in the new coordinates. It is easy to see that ${y = x,}\;\; {\Rightarrow y - x = 0,}\;\; {\Rightarrow u = 0,}$ ${y = 2x,}\;\; {\Rightarrow y - 2x = 0,}\;\; {\Rightarrow v = 0.}$
 Fig.3 Fig.4
We notice that ${u - v } = {\left( {y - x} \right) - \left( {y - 2x} \right) = x.}$ Hence, ${y = x + u } = {u - v + u } = {2u - v.}$ Then we have ${x + y = 2,}\;\; {\Rightarrow u - v + 2u - v = 2,}\;\; {\Rightarrow 3u - 2v = 2.}$ When $$v = 0,$$ we have $$u = {\large\frac{2}{3}\normalsize}.$$ And when $$u = 0,$$ then $$v = -1.$$ As a result, we can draw the pullback region $$S$$ (Figure $$4$$ above). It looks as a right triangle.

The equation of the line $$3u - 2v = 2$$ can be written as ${3u - 2v = 2,}\;\; {\Rightarrow v = \frac{{3u - 2}}{2} } = {\frac{3}{2}u - 1.}$ Find the Jacobian: ${\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\ {\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial \left( {u - v} \right)}}{{\partial u}}}&{\frac{{\partial \left( {u - v} \right)}}{{\partial v}}}\\ {\frac{{\partial \left( {2u - v} \right)}}{{\partial u}}}&{\frac{{\partial \left( {2u - v} \right)}}{{\partial v}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} 1&{ - 1}\\ 2&{ - 1} \end{array}} \right| } = {1 \cdot \left( { - 1} \right) - \left( { - 1} \right) \cdot 2 = 1.}$ Hence, $$dxdy = dudv$$ and the initial double integral is ${\iint\limits_R {\left( {x + y} \right)dxdy} } = {\iint\limits_S {\left( {u - v + 2u - v} \right)dudv} } = {\iint\limits_S {\left( {3u - 2v} \right)dudv} } = {\int\limits_0^{\frac{2}{3}} {\left[ {\int\limits_{\frac{3}{2}u - 1}^0 {\left( {3u - 2v} \right)dv} } \right]du} } = {\int\limits_0^{\frac{2}{3}} {\left[ {\left. {\left( {3uv - {v^2}} \right)} \right|_{v = \frac{3}{2}u - 1}^0} \right]du} } = { - \int\limits_0^{\frac{2}{3}} {\left[ {3u\left( {\frac{3}{2}u - 1} \right) - {{\left( {\frac{3}{2}u - 1} \right)}^2}} \right]du} } = { - \int\limits_0^{\frac{2}{3}} {\left( {\frac{{9{u^2}}}{2} - 3u - \frac{{9{u^2}}}{4} + 3u - 1} \right)du} } = { - \int\limits_0^{\frac{2}{3}} {\left( {\frac{{9{u^2}}}{4} - 1} \right)du} } = {\left. {\left( {u - \frac{9}{4}\frac{{{u^3}}}{3}} \right)} \right|_0^{\frac{2}{3}} } = {\frac{2}{3} - \frac{3}{4} \cdot {\left( {\frac{2}{3}} \right)^3} = \frac{4}{9}.}$
Example 3
Calculate the double integral $$\iint\limits_R {dxdy},$$ where the region $$R$$ is bounded by the parabolas $${y^2} = 2x,$$ $${y^2} = 3x$$ and hyperbolas $$xy = 1,$$ $$xy = 2.$$

Solution.
The region $$R$$ is sketched in Figure $$5.$$
 Fig.5
We apply the following substitution of variables to simplify the region $$R:$$ ${\left\{ \begin{array}{l} u = \frac{{{y^2}}}{x}\\ v = xy \end{array} \right..}$ The pulback $$S$$ of the region $$R$$ is defined as follows ${{y^2} = 2x,}\;\; {\Rightarrow \frac{{{y^2}}}{x} = 2,}\;\; {\Rightarrow u = 2,}$ ${{y^2} = 3x,}\;\; {\Rightarrow \frac{{{y^2}}}{x} = 3,}\;\; {\Rightarrow u = 3,}$ $xy = 1,\;\; \Rightarrow v = 1,$ $xy = 2,\;\; \Rightarrow v = 2.$ As can be seen, the region $$S$$ is the rectangle. To find the Jacobian of the transformation, we express the variables $$x, y$$ in terms of $$u, v$$. $u = \frac{{{y^2}}}{x},\;\; \Rightarrow x = \frac{{{y^2}}}{u},$ ${v = xy,}\;\; {\Rightarrow v = \frac{{{y^2}}}{u} \cdot y,}\;\; {\Rightarrow {y^3} = uv.}$ Then $y = \sqrt[3]{{uv}} = {u^{\frac{1}{3}}}{v^{\frac{1}{3}}},$ ${x = \frac{{{y^2}}}{u} } = {\frac{{\sqrt[3]{{{u^2}{v^2}}}}}{u} } = {\sqrt[3]{{\frac{{{v^2}}}{u}}} } = {{u^{ - \frac{1}{3}}}{v^{\frac{2}{3}}}.}$ Find the Jacobian: ${\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\ {\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial \left( {{u^{ - \frac{1}{3}}}{v^{\frac{2}{3}}}} \right)}}{{\partial u}}}&{\frac{{\partial \left( {{u^{ - \frac{1}{3}}}{v^{\frac{2}{3}}}} \right)}}{{\partial v}}}\\ {\frac{{\partial \left( {{u^{\frac{1}{3}}}{v^{\frac{1}{3}}}} \right)}}{{\partial u}}}&{\frac{{\partial \left( {{u^{\frac{1}{3}}}{v^{\frac{1}{3}}}} \right)}}{{\partial v}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {{v^{\frac{2}{3}}}\left( { - \frac{1}{3}{u^{ - \frac{4}{3}}}} \right)}&{{u^{ - \frac{1}{3}}} \cdot \frac{2}{3}{v^{ - \frac{1}{3}}}}\\ {\frac{1}{3}{u^{ - \frac{2}{3}}}{v^{\frac{1}{3}}}}&{\frac{1}{3}{v^{ - \frac{2}{3}}}{u^{\frac{1}{3}}}} \end{array}} \right| } = { - \frac{1}{3}{u^{ - \frac{4}{3}}}{v^{\frac{2}{3}}} \cdot \frac{1}{3}{u^{\frac{1}{3}}}{v^{ - \frac{2}{3}}} } - {\frac{2}{3}{u^{ - \frac{1}{3}}}{v^{ - \frac{1}{3}}} \cdot \frac{1}{3}{u^{ - \frac{2}{3}}}{v^{\frac{1}{3}}} } = { - \frac{1}{9}{u^{ - 1}} - \frac{2}{9}{u^{ - 1}} } = { - \frac{1}{3}{u^{ - 1}} = - \frac{1}{{3u}}.}$ The relation between the differentials is ${dxdy = \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv } = {\left| { - \frac{1}{{3u}}} \right|dudv } = {\frac{{dudv}}{{3u}}.}$ Then we can write the integral as ${\iint\limits_R {dxdy} } = {\iint\limits_S {\frac{{dudv}}{{3u}}} } = {\int\limits_2^3 {\frac{{du}}{{3u}}} \int\limits_1^2 {dv} } = {\frac{1}{3}\left. {\left( {\ln u} \right)} \right|_2^3 \cdot \left. v \right|_2^3 } = {\frac{1}{3}\left( {\ln 3 - \ln 2} \right) \cdot \left( {2 - 1} \right) } = {\frac{1}{3}\ln \frac{3}{2}.}$
Example 4
Evaluate the integral $$\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy},$$ where $$R$$ is bounded by the lines $$y = x,$$ $$y = x + a,$$ $$y = a,$$ $$y = 2a\;\left(a > 0\right).$$

Solution.
The region of integration $$R$$ is a parallelogram and shown in Figure $$6.$$
 Fig.6 Fig.7
We can make the following change of variables: ${\left\{ \begin{array}{l} u = y - x\\ v = y \end{array} \right.}\;\; {\text{or}\;\;\left\{ \begin{array}{l} x = y - u = v - u\\ y = v \end{array} \right..}$ The purpose of this change is to simplify shape of the region of integration $$R.$$
The image $$S$$ of $$R$$ in terms of $$\left( {u,v} \right)$$ is defined as ${y = x,}\;\; {\Rightarrow y - x = 0,}\;\; {\Rightarrow u = 0,}$ ${y = x + a,}\;\; {\Rightarrow y - x = a,}\;\; {\Rightarrow u = a,}$ $y = a,\;\; \Rightarrow v = a,$ $y = 2a,\;\; \Rightarrow v = 2a.$ As seen from the Figure $$7,$$ $$S$$ is the rectangular region. Calculate the Jacobian. ${\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\ {\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial \left( {v - u} \right)}}{{\partial u}}}&{\frac{{\partial \left( {v - u} \right)}}{{\partial v}}}\\ {\frac{{\partial v}}{{\partial u}}}&{\frac{{\partial v}}{{\partial v}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} { - 1}&1\\ 0&1 \end{array}} \right| } = { - 1 \cdot 1 - 1 \cdot 0 = - 1,}$ so that ${dxdy = \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv } = {\left| { - 1} \right| \cdot dudv = dudv.}$ Now we can write the double integral as ${\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy} } = {\iint\limits_S {\left[ {{{\left( {v - u} \right)}^2} + {v^2}} \right]dudv} } = {\iint\limits_S {\left( {{v^2} - 2uv + {u^2} + {v^2}} \right)dudv} } = {\int\limits_a^{2a} {\left[ {\int\limits_0^a {\left( {2{v^2} - 2uv + {u^2}} \right)du} } \right]dv} } = {\int\limits_a^{2a} {\left[ {\left. {\left( {2{v^2}u - v{u^2} + \frac{{{u^3}}}{3}} \right)} \right|_{u = 0}^a} \right]dv} } = {\int\limits_a^{2a} {\left( {2a{v^2} - {a^2}v + \frac{{{a^3}}}{3}} \right)dv} } = {\left. {\left( {2a \cdot \frac{{{v^3}}}{3} - {a^2} \cdot \frac{{{v^2}}}{2} + \frac{{{a^3}}}{3} \cdot v} \right)} \right|_a^{2a} } = {\left( {\frac{{2a}}{3} \cdot 8{a^3} - \frac{{{a^2}}}{2} \cdot 4{a^2} + \frac{{{a^3}}}{3} \cdot 2a} \right) } - {\left( {\frac{{2a}}{3} \cdot {a^3} - \frac{{{a^2}}}{2} \cdot {a^2} + \frac{{{a^3}}}{3} \cdot a} \right) } = {\frac{{7{a^4}}}{2}.}$