### IIT JAM MATHS

DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
LINE INTEGRAL
GREENS THEOREM
SURFACE INTEGRAL
GAUSS DIVERGENCE THEOREM
STOKES THEOREM
CONSERVATIVE VECTOR FIELD
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-4

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Example 1 :

Change the order of integration. $I = \int\limits_0^{2a} {\int\limits_{\sqrt {2ax - {x^2}} }^{\sqrt {2ax} } {fd} ydx}$

Solution :

The given integral is

$I = \int\limits_0^{2a} {\int\limits_{\sqrt {2ax - {x^2}} }^{\sqrt {2ax} } {fd} ydx}$

$I = \int\limits_{x = a}^{x = b} {\int\limits_{y = {g_1}(x)}^{y = {g_2}(x)} {fdydx}}$

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

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

${C_2}:y = \sqrt {2ax}$

${C_3}:x = 0$

${C_4}:x = 2a$

Let us trace all the curves to get region of integration

After changing the order of integration, the integral changes to

$I =\int\limits_{y = c}^{y=d} {\int\limits_{x={g_1}(y)}^{x = {g_2}(y)}{fd} xdy}$

The domain has to be partitioned by line y = a into three regions as shown in figure.

For region I, x varies from $x = \dfrac{{{y^2}}}{{2a}}$ to $x = a - \sqrt {{a^2} - {y^2}}$ and y varies from y = 0 to y = a

For region II, x varies from $x = a + \sqrt {{a^2} - {y^2}}$ to x = 2a and y varies from y = 0 to y = a

For region III, x varies from $x = \dfrac{{{y^2}}}{{2a}}$ to x = 2a and y varies from y = a to y = 2a

So, after changing the order, the integral becomes

$I = \int\limits_0^a {\int\limits_{{y^2}/2a}^{a - \sqrt {{a^2} - {y^2}} } {fdxdy} } + \int\limits_0^a {\int\limits_{a + \sqrt {{a^2} - {y^2}} }^{2a} {fdxdy} } + \int\limits_a^{2a} {\int\limits_{{y^2}/2a}^{2a} {fdxdy} }$