### IIT JAM MATHS

DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
GRADIENT, DIVERGENCE AND CURL
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-1

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Let us take the following example :

Theorem 1 :

Change the order of integration in $\int\limits_0^{a/2} {\int\limits_{{x^2}/a}^{x - {x^2}/a} {fdydx} }$ where f is a function of x and y.

Solution :

Step 1 : The first step involves getting the bounding curves of domain of integration.

Let us compare the limits of given integral to its standard form $\int\limits_0^{a/2} {\int\limits_{{x^2}/a}^{x - {x^2}/a} {fdydx}} \equiv \int\limits_{x = a}^{x = b} {\int\limits_{y = f(x)}^{y = {f_2}(x)} {fdydx}}$ comparing the limit, we get the bounding curves

The bounding curves are

$C_1 : y = \dfrac{{{x^2}}}{a}$

$C_2 : y = x - \dfrac{{{x^2}}}{a}$

$C_4 : x = a/2$

The first two curves are the parabolas

$C_1 : x^2 = ay$

and last two are straight lines

$C_2 : (x - a/2)^2 = -a(y - a/4)$ $C_3 : x = 0$ $C_4 : x = a/2$

Step 2 : The second step involves the plotting of the bounding curves and getting the domain of integration.
If the equation of curve is quadratic in one variable and linear in other variable, it will be a parabolas.

The equation of parabola is usually given in any of the two forms
(a) $(y - \beta)^2 = 4a(x - \alpha)$

(b) $(x - \alpha)^2 = 4b(y - \beta)$

The curves $C_1, C_2, C_3, C_4$ are plotted as

The region bounded by all the four curves is the domain of integration.

Step 3 : Third step involves determining the limits after changing the order.
On changing the order, the integral converts to

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

For finding the limit of x, let us draw an arrow parallel to x-axis and intersecting the region of integration arbitrarily

The arrow first intersects curve $C_2$ at point A, write the equation of curve $C_2$ in the form of $x = g_1(y)$

$x = \dfrac{a}{2} \pm \dfrac{1}{2}\sqrt {{a^2} - 4ay}$

Since, point A lies towards positive x

So, $x = \dfrac{a}{2} + \dfrac{1}{2}\sqrt {{a^2} - 4ay}$

This is the lower limit of x

The arrow then intersects curve $C_1$ at point B, write the equation of $C_1$ in the form of $x = g_2(y)$

$x = \sqrt {ay}$

This is the upper limit of x.

In order to transverse the whole region of integration the arrow has to be moved from y = 0 to $y = \dfrac{a}{4}$ so, these are the limit of y.
Hence, the integral after changing the order becomes

$I = \int\limits_0^{a/4} {\int\limits_{\dfrac{a}{2} + \dfrac{1}{2}\sqrt {{a^2} - 4ay} }^{\sqrt {ay} } {fdxdy} }$