Double Integrals over General Regions

If a region R of type I is bounded by x = a, x = b, y = p(x) and y = q(x) with a < b and p(x) < q(x) for all x ∈ [a, b], then using the Fubini's theorem the double integral can be written as

\[\iint\limits_R {f\left( {x,y} \right)dxdy} = \int\limits_{x = a}^{x = b} {\int\limits_{y = p\left( x \right)}^{y = q\left( x \right)} {f\left( {x,y} \right)dydx} } .\]

The similar formula exists for regions of type \(II.\) If a region \(R\) of type \(II\) is bounded by the graphs of the functions \(x = u\left( y \right),\) \(x = v\left( y \right),\) \(y = c,\) \(y = d\) provided that \(c \lt d\) and \(u\left( y \right) \lt v\left( y \right)\) for all \(y \in \left[ {c,d} \right],\) then the double integral over the region \(R\) is expressed through the iterated integral by the Fubini's theorem

\[\iint\limits_R {f\left( {x,y} \right)dxdy} = \int\limits_{y = c}^{y = d} {\int\limits_{x = u\left( y \right)}^{x = v\left( y \right)} {f\left( {x,y} \right)dxdy} } .\]

It is sometimes useful to break the region \(R\) up into two or more smaller regions, and integrate over each separately.

Solved Problems

Example 1.

Calculate the double integral \[\iint\limits_R {\left( {x - y} \right)dxdy}.\] The region of integration \(R\) is bounded

\[x = 0, x = 1, y = x,y = 2 - {x^2}.\]

Solution.

We can represent the region \(R\) in the form \({R =}\) \({ \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,\;}\right.}\) \({\left.{ x \le y \le 2 - {x^2}} \right\}.}\)

\[R = \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,\; x \le y \le 2 - {x^2}} \right\}.\]

So \(R\) is the region of type \(I\) (see Figure \(1\)).

A region of integration bounded by straight lines and a parabola — Figure 1.

According to the Fubini's formula,

\[\iint\limits_R {\left( {x - y} \right)dxdy} = \int\limits_0^1 {\int\limits_x^{2 - {x^2}} {\left( {x - y} \right)dydx} } = \int\limits_0^1 {\left[ {\int\limits_x^{2 - {x^2}} {\left( {x - y} \right)dy} } \right]dx} .\]

Calculate first the inner integral:

\[\int\limits_x^{2 - {x^2}} {\left( {x - y} \right)dy} = \left. {\left( {xy - \frac{{{y^2}}}{2}} \right)} \right|_{y = x}^{2 - {x^2}} = [ {x\left( {2 - {x^2}} \right) - \frac{{{{\left( {2 - {x^2}} \right)}^2}}}{2}}] - \left[ {{x^2} - \frac{{{x^2}}}{2}} \right] = - \frac{{{x^4}}}{2} - {x^3} + \frac{{3{x^2}}}{2} + 2x - 2.\]

Now we can compute the outer integral:

\[\int\limits_0^1 {\left( { - \frac{{{x^4}}}{2} - {x^3} + \frac{{3{x^2}}}{2} + 2x - 2} \right)dx} = \left. {\left( { - \frac{{{x^5}}}{{10}} - \frac{{{x^4}}}{4} + \frac{{{x^3}}}{2} + {x^2} - 2x} \right)} \right|_0^1 = - \frac{{17}}{{20}}.\]

Example 2.

Calculate the integral \[\iint\limits_R {\left( {x + y} \right)dxdy}.\] The region of integration \(R\) is bounded by the lines

\[x = 0, y = 0, x + y = 2.\]

Solution.

We can represent the region \(R\) as the set

\[R = \left\{ {\left( {x,y} \right)|\;0 \le x \le 2,\; 0 \le y \le 2 - x} \right\}.\]

A region of integration bounded by a straight line and coordinate axes — Figure 2.

The region \(R\) belongs to type \(I.\) Converting the double integral into the iterated one, we get

\[ \iint\limits_R {\left( {x + y} \right)dxdy} = \int\limits_0^2 {\int\limits_0^{2 - x} {\left( {x + y} \right)dydx} } = \int\limits_0^2 {\left[ {\int\limits_0^{2 - x} {\left( {x + y} \right)dy} } \right]dx} = \int\limits_0^2 {\left[ {\left. {\left( {xy + \frac{{{y^2}}}{2}} \right)} \right|_{y = 0}^{2 - x}} \right]dx} = \int\limits_0^2 {\left[ {x\left( {2 - x} \right) + \frac{{{{\left( {2 - x} \right)}^2}}}{2}} \right]dx} = \int\limits_0^2 {\left( {2 - \frac{{{x^2}}}{2}} \right)dx} = \left. {\left( {2x - \frac{{{x^3}}}{6}} \right)} \right|_0^2 = \frac{8}{3}.\]

Example 3.

Evaluate the integral \[\iint\limits_R {xdxdy},\] where the region of integration \(R\) is bounded by the graphs of the functions

\[y = {x^3}, x + y = 2, x = 0.\]

Solution.

The region of integration \(R\) is shown in Figure \(3.\)

Region of integration bounded by the graphs of the functions y=x^3, x+y=2, x=0. — Figure 3.

The curve \(y = {x^3}\) and the line \(x + y = 2\) intersect at \(\left( {1,1} \right).\) Then, by Fubini's theorem, the double integral is

\[\iint\limits_R {xdxdy} = \int\limits_0^1 {\int\limits_{{x^3}}^{2 - x} {xdydx} } = \int\limits_0^1 {\left[ {\int\limits_{{x^3}}^{2 - x} {xdy} } \right]dx} = \int\limits_0^1 {\left[ {\left. {\left( {xy} \right)} \right|_{y = {x^3}}^{2 - x}} \right]dx} = \int\limits_0^1 {\left[ {x\left( {2 - x} \right) - {x^4}} \right]dx} = \left. {\left( {{x^2} - \frac{{{x^3}}}{3} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = 1 - \frac{1}{3} - \frac{1}{5} = \frac{7}{{15}}.\]

Example 4.

Find the integral \[\iint\limits_R {{x^2}ydxdy},\] where the region \(R\) is the segment of a circle. The boundaries of the segment are defined by the equations

\[{x^2} + {y^2} = 4, x + y - 2 = 0.\]

Solution.

The circle \({x^2} + {y^2} = 4\) has the radius \(2\) and centre at the origin (Figure \(4\)).

Region of integration bounded by a circle and a straight line. — Figure 4.

Since the upper half of the circle is equivalent to \(y = \sqrt {4 - {x^2}},\) the double integral can be written in the following form:

\[ \iint\limits_R {{x^2}ydxdy} = \int\limits_0^2 {\int\limits_{2 - x}^{\sqrt {4 - {x^2}} } {{x^2}ydxdy} } = \int\limits_0^2 {\left[ {\int\limits_{2 - x}^{\sqrt {4 - {x^2}} } {{x^2}ydy} } \right]dx} = \int\limits_0^2 {\left[ {\left. {\left( {\frac{{{x^2}{y^2}}}{2}} \right)} \right|_{y = 2 - x}^{\sqrt {4 - {x^2}} }} \right]dx} = \frac{1}{2}\int\limits_0^2 {\left[ {{x^2}\left( {4 - {x^2}} \right) - {x^2}{{\left( {2 - x} \right)}^2}} \right]dx} = \frac{1}{2}\int\limits_0^2 {\left[ {4{x^2} - {x^4} - {x^2}\left( {4 - 4x + {x^2}} \right)} \right]dx} = \frac{1}{2}\int\limits_0^2 {\left( {4{x^3} - 2{x^4}} \right)dx} = \frac{1}{2}\left. {\left( {{x^4} - \frac{{2{x^5}}}{5}} \right)} \right|_0^2 = \frac{1}{2}\left( {16 - \frac{{64}}{5}} \right) = \frac{8}{5}.\]

See more problems on Page 2.