# Change of Order-6

Example 1 :

Evaluate $\int \int \cos (x + y) dx dy$ over the region enclosed by $x = 0, y = \pi$ and $y = x$.

Solution :

The region of integration is shown in the figure

$I = \int \int \cos (x + y)dx dy$

$= \int_{0}^{\pi} \int_{0}^{y} \cos(x + y) dx dy$

$= \int_{0}^{\pi} \left [ \sin (x + y) \right ]_{0}^{y} dy$

$= \int_{0}^{\pi} \left ( \sin 2y - \sin y \right ) dy$

$= - \dfrac{1}{2} \left . \cos 2y + \cos y \right |_{0}^{\pi}$

$= -2$

Example 2 :

Evaluate the integral $\int \int xe^{x^{2} - y^{2}} dx dy$ over the region bounded by the lines y = x, y = x – 1, y = 0 and y = 1.

Solution :

The region of integration is shown in figure. It is easier to evaluate the integral first w.r.t. x and then w.r.t. y.

$I = \int_{0}^{1} \int_{y}^{y + 1} xe^{x^{2} - y^{2}} dx dy$

$= \dfrac{1}{2} \int_{0}^{1} e^{-y^{2}} \left [ e^{x^{2}} \right ]_{y}^{y + 1} dy$

$= \dfrac{1}{2} \int_{0}^{1} \left (e^{2y + 1} - 1 \right ) dy$

$= \dfrac{1}{2} \left [ \dfrac{e}{2} e^{2y} - y \right ]_{0}^{1}$

$= \dfrac{1}{4} [e^{3} - e - 2]$

Example 3 :

Evaluate $\int \int (x^{2} + y^{2}) dx dy$ over the region bounded by xy = 1, y = 0, y = x and x = 2.

Solution :

The region of integration is as shown in figure. It has to be partitioned by line x = 1, before integration first w.r.t. y and then w.r.t x,

$I = \int \int (x^{2} + y^{2})dy dx$

$= \int \int (x^{2} + y^{2}) dy dx + \int \int (x^{2} + y^{2})dy dx = I_{1} + I_{2}$

$I_{1} = \int_{0}^{1} \int_{0}^{x} (x^{2} + y^{2})dy dx$

$= \int_{0}^{1} \left [ x^{2} y + \dfrac{y^{3}}{3} \right ]_{0}^{x} dx$

$= \int_{0}^{1} \dfrac{4}{3} x^{3} dx$

$= \dfrac{4}{3} \left [ \dfrac{x^{4}}{4} \right ]_{0}^{1} = \dfrac{1}{3}$
$I_{2} = \int_{1}^{2} \int_{0}^{1/x} (x^{2} + y^{2}) dy dx$

$= \int_{1}^{2} x^{2} y + \left . \dfrac{y^{3}}{3} \right |_{0}^{1/x} dx$

$= \int_{1}^{2} \left ( x + \dfrac{1}{3x^{3}} \right ) dx$

$= \left [ \dfrac{x^{2}}{2} - \dfrac{1}{6x^{{2}}} \right ]_{1}^{2}$

$= \dfrac{39}{24}$

So, $I = I_{1} + I_{2}$

$= \dfrac{47}{24}$