Calculus

Double Integrals

Change of Variables in Double Integrals

Page 1
Problem 1
Page 2
Problems 2-4

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

\[
{\iint\limits_R {f\left( {x,y} \right)dxdy} }
= {\iint\limits_S {f\left[ {x\left( {u,v} \right),y\left( {u,v} \right)} \right] }\kern0pt{ \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dxdy} ,}
\]

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.

Solved Problems

Click on problem description to see solution.

 Example 1

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

 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 \(y = x,\) \(y = 2x,\) \(x + y = 2.\)

 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 \(xy = 1,\) \(xy = 2.\)

 Example 4

Evaluate the integral \(\iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy},\) where \(R\) is bounded by the lines \(y = x,\) \(y = x + a,\) \(y = a,\) \(y = 2a\) \(\left(a \gt 0\right).\)

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 \(2.\)

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

Region of integration bounded by the straight lines y=x+1, y=x-3, y=-x/3+2, y=-x/3+4.

Figure 1.

The pullback of the regoin of integration R

Figure 2.

\[
{\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| }
= {\left| {{{\left( {\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}} \right)}^{ – 1}}} \right| }
= {\left| {\frac{1}{{ – \frac{4}{3}}}} \right| }={ \frac{3}{4}.}
\] Hence, the differential is
\[
{dxdy }={ \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv }
= {\frac{3}{4}dudv.}
\] As it can be seen, calculating the integral in the new variables \(\left( {u,v} \right)\) is much simpler:
\[
{\iint\limits_R {\left( {y – x} \right)dxdy} }
= {\iint\limits_S {\left( u \cdot \frac{3}{4}dudv \right)} }
= {\frac{3}{4}\int\limits_{ – 3}^1 {udu} \int\limits_2^4 {dv} }
= {\frac{3}{4}\left. {\left( {\frac{{{u^2}}}{2}} \right)} \right|_{ – 3}^1 \cdot \left. v \right|_2^4 }
= {\frac{3}{4}\left( {\frac{1}{2} – \frac{9}{2}} \right) \cdot \left( {4 – 2} \right) }={ – 6.}
\]

Page 1
Problem 1
Page 2
Problems 2-4