### 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-3

Example 1 :

Change the order of double integration $\int\limits_{ - 1/\sqrt 2 }^{1/\sqrt 2 } {\int\limits_{ - \sqrt {1 - 2{x^2}} }^{\sqrt {1 - 2{x^2}} } {fdy\,dx} }$

Solution :

The given integral is

$I = \int\limits_{ - 1/\sqrt 2 }^{1/\sqrt 2 } {\int\limits_{ - \sqrt {1 - 2{x^2}} }^{\sqrt {1 - 2{x^2}} } {fdy\,dx} }$

Comparing the two integral, the curves bounding the region of integration are given by

${C_1}:y = - \sqrt {1 - 2{x^2}}$

${C_2}:y = \sqrt {1 - 2{x^2}}$

${C_3}: x = - \dfrac{1}{{\sqrt 2 }}$

${C_4}: x = \dfrac{1}{{\sqrt 2 }}$

$C_1$ & $C_2$ are the part of ellipse

$2x^2 + y^2 = 1$

The region of integration is shown in figure

After changing the order, the given integral converts to

$I = \int\limits_{y = c}^{y = d} {\int\limits_{x = {g_1}(y)}^{x = {g_2}(y)} {fdx\,dy} }$

To get the limit of x, let us draw an arrow parallel to x-axis intersecting the region of integration arbitrarily.

The arrow enters the region at A and leaves at point B.

Let us write the equation of curve at A and B in the form x = g(y)

At $A, x = - \sqrt {\dfrac{{1 - {y^2}}}{2}}$ and at $B, x = \sqrt {\dfrac{{1 - {y^2}}}{2}}$

The arrow has to be moved from y = –1 to y = 1 so as the traverse the whole region of integration.

Hence, the unit of x is from $- \sqrt {\dfrac{{1 - {y^2}}}{2}}$ to $\sqrt {\dfrac{{1 - {y^2}}}{2}}$ and that of y from –1 to +1.

So, after changing the order

$I = \int_{ - 1}^1 \int_{ \sqrt{\dfrac{1-y^2}{2}}}^{\sqrt{\dfrac{1-y^2}{2}}} fdxdy$

Example 2 :

Change the order of double integration. $I = \int\limits_1^2 {\left( {\int\limits_x^{2x} {fdy} } \right)dx}$

Solution :

The given integral is

$I = \int\limits_1^2 {\left( {\int\limits_x^{2x} {fdy} } \right)dx}$

$\int\limits_{x = a}^{x = b} {\left( {\int\limits_{y = {f_1}(x)}^{y = {f_2}(x)} {fdy} } \right)dx}$

Comparing the two integral, the curves bounding the region of integration are given by

$C_1 : y = x$

$C_2 : y = 2x$

$C_3 : x = 1$

$C_4 : x = 2$

Let us draw these curves to get the region of integration.
After changing the order, the given integral converts to

$I = \int\limits_{y = c}^{y=d}{\int\limits_{x = g(y)}^{x = {g_2}(y)}{fdxdy}}$

To get the limit of x, let us draw an arrow parallel to x-axis intersecting the region of integration arbitrarily.
From the region of integration plotted in figure we find that if arrow is drawn below line, y = 2, the curves intersected are x = 1 and y = x but if arrow is drawn above line y = 2 the curves intersected are y = 2x and x = 2.
In this case the region of integration has to be partitioned into region I and region II by partition line y = 2.
So, the given integral will be broken into two integrals corresponding to I and II while changing the order. For region I, the limit of x varies from x = 1 to x = y and that of y varies from y = 1 to y = 2.
For region II, the limit of x varies from $x = \dfrac{y}{2}$ to x = 2 and that of y varies from y = 2 to y = 4.
So, after changing the order, the integral changes to

$I = \int\limits_1^2 {\int\limits_1^y {fdxdy} } + \int\limits_2^4 {\int\limits_{y/2}^2 {fdxdy} }$