# Double Integrals over General Regions

## Solved Problems

Click or tap a problem to see the solution.

### Example 5

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

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

### Example 6

Find the integral $\iint\limits_R {ydydx},$ where R is bounded by the straight line y = 2x and parabola y = 3 − x².

### Example 7

Calculate the double integral $\iint\limits_R {x\sin ydydx},$ where $$R$$ is bounded by

$y = 0, y = {x^2}, x = 1.$

### 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 $$C\left( {1,1} \right).$$

### 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.

### 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.

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 $$6.$$

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 $$R = \left\{ {\left( {x,y} \right)|\;0 \le x \le 1,}\right.$$ $$\left.{0 \le y \le {x^2}} \right\}$$ $$\left({\text{Figure }7}\right).$$

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 $$8.$$

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 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

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