# Double Integrals over Rectangular Regions

## Solved Problems

### Example 3.

Calculate the integral $\iint\limits_R {\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}}$ over the rectangular region

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

Solution.

Using the Fubini's theorem and expressing the double integral through the iterated integral (where the inner integral depends on $$x$$), we have

$\iint\limits_R {\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}} = \int\limits_1^2 {\int\limits_0^2 {\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}} } = \int\limits_1^2 {\left[ {\int\limits_0^2 {\frac{{dx}}{{{{\left( {x + y} \right)}^2}}}} } \right]dy} = \int\limits_1^2 {\left[ {\left. {\left( { - \frac{1}{{x + y}}} \right)} \right|_{x = 0}^2} \right]dy} = \int\limits_1^2 {\left[ { - \frac{1}{{y + 2}} + \frac{1}{y}} \right]dy} = \left. {\left[ {\ln y - \ln \left( {y + 2} \right)} \right]} \right|_1^2 = \left. {\left( {\ln \frac{y}{{y + 2}}} \right)} \right|_1^2 = \ln \frac{2}{4} - \ln \frac{1}{3} = \ln \frac{3}{2}.$

### Example 4.

Evaluate the integral $\iint\limits_R {\cos \left( {x + y} \right)dxdy}$ over the region

$R = \left\{ {\left( {x,y} \right)|\;0 \le x \le {\frac{\pi }{4}},\; 0 \le y \le {\frac{\pi }{4}}} \right\}.$

Solution.

Here we integrate first by $$x$$ and then by $$y:$$

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

### Example 5.

Evaluate the integral $\iint\limits_R {\left( {x - {y^2}} \right)dxdy}$ over the region

$R = \left\{ {\left( {x,y} \right)|\;2 \le x \le 3,\; 1 \le y \le 2} \right\}.$

Solution.

We express the double integral in terms of iterated integral. First we integrate by $$x,$$ then by $$y.$$

$\iint\limits_R {\left( {x - {y^2}} \right)dxdy} = \int\limits_1^2 {\int\limits_2^3 {\left( {x - {y^2}} \right)dxdy} } = \int\limits_1^2 {\left[ {\int\limits_2^3 {\left( {x - {y^2}} \right)dx} } \right]dy} = \int\limits_1^2 {\left[ {\left. {\left( {\frac{{{x^2}}}{2} - {y^2}x} \right)} \right|_{x = 2}^3} \right]dy} = \int\limits_1^2 {\Big[ {\left( {\frac{9}{2} - 3{y^2}} \right) - \left( {2 - 2{y^2}} \right)} \Big] dy} = \int\limits_1^2 {\left( {\frac{5}{2} - {y^2}} \right)dy} = \left. {\left( {\frac{5}{2}y - \frac{{{y^3}}}{3}} \right)} \right|_1^2 = \left( {5 - \frac{8}{3}} \right) - \left( {\frac{5}{2} - \frac{1}{3}} \right) = \frac{1}{6}.$

We can change the order of integration. Certainly, the result will be the same:

$\iint\limits_R {\left( {x - {y^2}} \right)dxdy} = \int\limits_2^3 {\int\limits_1^2 {\left( {x - {y^2}} \right)dydx} } = \int\limits_2^3 {\left[ {\int\limits_1^2 {\left( {x - {y^2}} \right)dy} } \right]dx} = \int\limits_2^3 {\left[ {\left. {\left( {xy - \frac{{{y^3}}}{3}} \right)} \right|_{y = 1}^2} \right]dx} = \int\limits_2^3 {\left[ {\left( {2x - \frac{8}{3}} \right) - \left( {x - \frac{1}{3}} \right)} \right]dx} = \int\limits_2^3 {\left( {x - \frac{7}{3}} \right)dx} = \left. {\left( {\frac{{{x^2}}}{2} - \frac{{7x}}{3}} \right)} \right|_2^3 = \left( {\frac{9}{2} - 7} \right) - \left( {2 - \frac{{14}}{3}} \right) = \frac{1}{6}.$