Double Integrals over General Regions

Solved Problems

Example 5.

Find the double integral \[\iint\limits_R {\sin \left( {x + y} \right)dxdy},\] defined in the region \(R\) bounded by the lines

\[y = x, x + y = {\frac{\pi }{2}}, y = 0.\]

Solution.

Region of integration bounded by the lines y=x, y+x=pi/2 and the x-axis. — Figure 5.

Considering the region of integration \(R\) as a type \(II\) region (Figure \(5\)), we can transform the double integral into iterated one and calculate it in the following way:

\[ \iint\limits_R {\sin \left( {x + y} \right)dxdy} = \int\limits_0^{\frac{\pi }{4}} {\left[ {\int\limits_0^{\frac{\pi }{2} - y} {\sin \left( {x + y} \right)dx} } \right]dy} = \int\limits_0^{\frac{\pi }{4}} {\left[ {\left. {\left( { - \cos \left( {x + y} \right)} \right)} \right|_{x = y}^{\frac{\pi }{2} - y}} \right]dy} = - \int\limits_0^{\frac{\pi }{4}} {\Big[ {\cos \left( {\frac{\pi }{2} - y + y} \right) - \cos 2y} \Big] dy} = \int\limits_0^{\frac{\pi }{4}} {\cos 2ydy} = \left. {\left( {\frac{{\sin 2y}}{2}} \right)} \right|_0^{\frac{\pi }{4}} = \frac{1}{2}\left( {\sin \frac{\pi }{2} - \sin 0} \right) = \frac{1}{2}.\]

Example 6.

Find the integral \[\iint\limits_R {ydydx},\] where \(R\) is bounded by the straight line \(y = 2x\) and parabola \(y = 3 - {x^2}.\)

Solution.

The region \(R\) is shown in Figure \(6.\)

Region of integration bounded by the parabola y=3-x^2 and straight line y=2x. — Figure 6.

Find the points of intersection of the line and parabola:

\[\left\{ \begin{array}{l} y = 2x\\ y = 3 - {x^2} \end{array} \right.,\;\; \Rightarrow 2x = 3 - {x^2},\;\; \Rightarrow {x^2} + 2x - 3 = 0,\;\;\Rightarrow {x_{1,2}} = \frac{{ - 2 \pm \sqrt {16} }}{2} = - 3;\;1.\]

Hence, the given lines bounding the region \(R\) intersect at the points \(\left( { - 3, - 6} \right)\) and \(\left( {1, 2} \right).\)

Then by Fubini's theorem, the integral is

\[\iint\limits_R {ydydx} = \int\limits_{ - 3}^1 {\left[ {\int\limits_{2x}^{3 - {x^2}} {ydy} } \right]dx} = \int\limits_{ - 3}^1 {\left[ {\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_{y = 2x}^{3 - {x^2}}} \right]dx} = \frac{1}{2}\int\limits_{ - 3}^1 {\left[ {{{\left( {3 - {x^2}} \right)}^2} - {{\left( {2x} \right)}^2}} \right]dx} = \frac{1}{2}\int\limits_{ - 3}^1 {\left( {9 - 10{x^2} + {x^4}} \right)dx} = \frac{1}{2}\left. {\left( {9x - \frac{{10{x^3}}}{3} + \frac{{{x^5}}}{5}} \right)} \right|_{ - 3}^1 = \frac{1}{2}\left[ {\left( {9 - \frac{{10}}{3} + \frac{1}{5}} \right) - \left( { - 27 + \frac{{10 \cdot 27}}{3} - \frac{{243}}{5}} \right)} \right] = - \frac{{64}}{{15}}.\]

Example 7.

Calculate the double integral \[\iint\limits_R {x\sin ydydx},\] where \(R\) is bounded by

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

Solution.

The region of integration \(R\) is defined by the set \(R = \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,}\right.\) \(\left.{0 \le y \le {x^2}} \right\}\) \(\left({\text{Figure }7}\right).\)

Region of integration bounded by the lines y=x^2, y=0, x=1. — Figure 7.

Applying Fubini's theorem, we obtain

\[I = \iint\limits_R {x\sin ydydx} = \int\limits_0^1 {\left[ {\int\limits_0^{{x^2}} {x\sin ydy} } \right]dx} = \int\limits_0^1 {\left[ {\left. {\left( { - x\cos y} \right)} \right|_{y = 0}^{{x^2}}} \right]dx} = \int\limits_0^1 {\left( { - x\cos {x^2} + x\cos 0} \right)dx} = \int\limits_0^1 {\left( {1 - \cos {x^2}} \right)xdx} .\]

To find the latter integral, we make the substitution:

\[z = {x^2},\;\;\Rightarrow dz = 2xdx,\;\; \Rightarrow xdx = \frac{{dz}}{2}.\]

When \(x = 0,\) we have \(x = 0.\) Hence at \(x = 1\) we have \(z = 1.\) Then the integral is given by

\[I = \int\limits_0^1 {\left( {1 - \cos {x^2}} \right)xdx} = \int\limits_0^1 {\left( {1 - \cos z} \right)\frac{{dz}}{2}} = \frac{1}{2}\int\limits_0^1 {\left( {1 - \cos z} \right)dz} = \frac{1}{2}\left. {\left( {z - \sin z} \right)} \right|_0^1 = \frac{{1 - \sin 1}}{2} \approx 0,08.\]

Example 8.

Calculate the integral \[\iint\limits_R {{e^x}dxdy}.\] The region of integration is the triangle with the vertices \(O\left( {0,0} \right),\) \(B\left( {0,1} \right),\) and \(C\left( {1,1} \right).\)

Solution.

The region \(R\) is shown in Figure \(8.\)

A triangular region of integration — Figure 8.

Obviously, that the equation of the side \(OC\) is \(y = x,\) and the equation of the side \(BC\) is \(y = 1.\)

Considering \(R\) as a type \(I\) region and using Fubini's theorem, we get

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

Now we apply integration by parts to compute the outer integral. Let \(u = 1 - x,\) \(dv = {e^x}dx.\) Then \(du = -dx,\) \(v = \int {{e^x}dx} = {e^x}.\) Hence,

\[I = \int\limits_0^1 {{e^x}\left( {1 - x} \right)dx} = \left. {\left[ {{e^x}\left( {1 - x} \right)} \right]} \right|_0^1 + \int\limits_0^1 {{e^x}dx} = \left. {\left[ {{e^x}\left( {1 - x} \right)} \right]} \right|_0^1 + \left. {\left[ {{e^x}} \right]} \right|_0^1 = \left. {\left[ {2{e^x} - x{e^x}} \right]} \right|_0^1 = 2e - e - 2 = e - 2.\]

Example 9.

Find the double integral \[\iint\limits_R {\left( {x + y} \right)dxdy},\] where the region \(R\) is a parallelogram with the sides

\[y = x, y = x + a, y = a, y = 2a,\]

\(a\) is a parameter.

Solution.

We will consider \(R\) as a region of type \(II\) \(\left({\text{Figure }9}\right).\)

The region of integration R is a parallelogram — Figure 9.

The values of \(y\) are those between \(a\) and \(2a\), while, given \(y,\) the relevant values of \(x\) are those between \(x = y - a\) and \(x = y.\) So the double integral is

\[ \iint\limits_R {\left( {x + y} \right)dxdy} = \int\limits_a^{2a} {\left[ {\int\limits_{y - a}^y {\left( {x + y} \right)dx} } \right]dy} = \int\limits_a^{2a} {\left[ {\left. {\left( {\frac{{{x^2}}}{2} + yx} \right)} \right|_{x = y - a}^y} \right]dy} = \int\limits_a^{2a} {\left[ {\left( {\frac{{{y^2}}}{2} + {y^2}} \right) - \left( {\frac{{{{\left( {y - a} \right)}^2}}}{2} + y\left( {y - a} \right)} \right)} \right]dy} = \int\limits_a^{2a} {\Big( {\frac{{3{y^2}}}{2} - \frac{{{y^2} - 2ay + {a^2}}}{2} - {y^2} + ay} \Big) dy\,} = \int\limits_a^{2a} {\left( {2ay - \frac{{{a^2}}}{2}} \right)dy} = \left. {\left( {\frac{{2a{y^2}}}{2} - \frac{{{a^2}y}}{2}} \right)} \right|_a^{2a} = \left. {\left( {a{y^2} - \frac{{{a^2}y}}{2}} \right)} \right|_a^{2a} = \left( {a \cdot {{\left( {2a} \right)}^2} - \frac{{{a^2}}}{2} \cdot 2a} \right) - \left( {a \cdot {a^2} - \frac{{{a^2}}}{2} \cdot a} \right) = 4{a^3} - {a^3} - {a^3} + \frac{{{a^3}}}{2} = \frac{{5{a^3}}}{2}.\]