Calculus

Double Integrals

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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

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

where the expression

\[\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}} \end{array}} \right| \ne 0\]

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 \(\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|\) 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: \(dxdy = \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv;\)
  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

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Example 1

Calculate the double integral \[\iint\limits_R {\left( {y - x} \right)dxdy},\] where the region \(R\) is bounded by

\[y = x + 1, y = x - 3, y = - {\frac{x}{3}} + 2, y = - {\frac{x}{3}} + 4.\]

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 1.

Calculate the double integral \[\iint\limits_R {\left( {y - x} \right)dxdy},\] where the region \(R\) is bounded by

\[y = x + 1, y = x - 3, y = - {\frac{x}{3}} + 2, y = - {\frac{x}{3}} + 4.\]

Solution.

The region \(R\) is sketched in Figure \(1.\)

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

We use change of variables to simplify the integral. By letting \(u = y - x,\) \(v = y + {\frac{x}{3}},\) 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 \(2.\)

The pullback of the region of integration R
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

\[\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {{{\left( {\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}} \right)}^{ - 1}}} \right| = \left| {\frac{1}{{ - \frac{4}{3}}}} \right| = \frac{3}{4}.\]

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:

\[\iint\limits_R {\left( {y - x} \right)dxdy} = \iint\limits_S {\left( u \cdot \frac{3}{4}dudv \right)} = \frac{3}{4}\int\limits_{ - 3}^1 {udu} \int\limits_2^4 {dv} = \frac{3}{4}\left. {\left( {\frac{{{u^2}}}{2}} \right)} \right|_{ - 3}^1 \cdot \left. v \right|_2^4 = \frac{3}{4}\left( {\frac{1}{2} - \frac{9}{2}} \right) \cdot \left( {4 - 2} \right) = - 6.\]

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 is shown in Figure \(3.\)

Region of integration bounded by the straight lines y=x, y=2x, x+y=2.
Figure 3.
The pullback S of the region of integration R.
Figure 4.

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

\[y = x,\;\; \Rightarrow y - x = 0,\;\; \Rightarrow u = 0,\]
\[y = 2x,\;\; \Rightarrow y - 2x = 0,\;\; \Rightarrow v = 0.\]

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 \(u = {\frac{2}{3}}.\) 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

\[\iint\limits_R {\left( {x + y} \right)dxdy} = \iint\limits_S {\left( {u - v + 2u - v} \right)dudv} = \iint\limits_S {\left( {3u - 2v} \right)dudv} = \int\limits_0^{\frac{2}{3}} {\left[ {\int\limits_{\frac{3}{2}u - 1}^0 {\left( {3u - 2v} \right)dv} } \right]du} = \int\limits_0^{\frac{2}{3}} {\left[ {\left. {\left( {3uv - {v^2}} \right)} \right|_{v = \frac{3}{2}u - 1}^0} \right]du} = - \int\limits_0^{\frac{2}{3}} {\left[ {3u\left( {\frac{3}{2}u - 1} \right) - {{\left( {\frac{3}{2}u - 1} \right)}^2}} \right]du} = - \int\limits_0^{\frac{2}{3}} {\left( {\frac{{9{u^2}}}{2} - 3u - \frac{{9{u^2}}}{4} + 3u - 1} \right)du} = - \int\limits_0^{\frac{2}{3}} {\left( {\frac{{9{u^2}}}{4} - 1} \right)du} = \left. {\left( {u - \frac{9}{4}\frac{{{u^3}}}{3}} \right)} \right|_0^{\frac{2}{3}} = \frac{2}{3} - \frac{3}{4} \cdot {\left( {\frac{2}{3}} \right)^3} = \frac{4}{9}.\]

See more problems on Page 2.

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