### IIT JAM MATHS

DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
LIMITS
CONTINUITY
Continuity
DIFFERENTIABILITY
APPLICATION OF DERIVATIVES
MEAN VALUE THEOREM
Part B: FUNCTION OF TWO VARIABLES
INTRODUCTION
DIFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE
ORTHOGONAL TRAJECTORY
DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
CAUCHY EULER EQUATIONS
DIFFERENTIAL EQUATION OF SECOND ORDER
Section 5: LINEAR ALGEBRA
VECTOR SPACE AND LINEAR EQUATIONS
ORTHOGONALITY
EIGENVALUES AND EIGENVECTORS
LINEAR TRANSFORMATION
Section 6: GROUP THEORY
Unit Group
Subgroup
CYCLIC GROUP
PERMUTATION GROUP
ISOMORPHISM OF GROUPS
EXTRNAL DIRECT PRODUCT
HOMOMORPHISM OF GROUP
SET THEORY
Set Theory
REAL NUMBERS
TOPOLOGY ON REAL LINE
Open Sets
REAL SEQUENCES
INFINITE SERIES
POWER SERIES

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

$= \dfrac{47}{24}$