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
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
CYCLIC GROUP
PERMUTATION GROUP
ISOMORPHISM OF GROUPS
EXTRNAL DIRECT PRODUCT
HOMOMORPHISM OF GROUP
SET THEORY
REAL NUMBERS
TOPOLOGY ON REAL LINE
REAL SEQUENCES
INFINITE SERIES
POWER SERIES

Change of Order-2

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Example 1 :

Change the order of integration in  \int\limits_0^{2a} {\int\limits_{{x^2}/4a}^{3a - x} {f(x,\,\,y)dydx} }

Solution :

On comparing the limits of integral is given form to its standard form

 \int\limits_0^{2a} {\int\limits_{{x^2}/4a}^{3a - x} {f(x,\,\,y)dydx} } = \int\limits_{x = a}^{x = b} {\int\limits_{y = {f_1}(x)}^{y = {f_2}(x)} {f(x,\,\,y)dydx} }

The bounding curves are given by

 C_1 : y = \dfrac{{{x^2}}}{{4a}} \Rightarrow x^2 = 4ay

 C_2 : y = 3a - y \Rightarrow x + y = 3a

 C_3 : x = 0

 C_4 : x = 2a

Let us plot all these curves

The domain of integration is the region bounded by all the four bounding curves as shown in figure.
The arrow drawn parallel to x-axis has two possibilities :

(a) If drawn below the line y = a, the arrow intersects y-axis and the parabola  x^2 = 4ay

(b) If drawn above the line y = a, the arrow intersects y-axis and a straight line x + y = 3a
In this case, we have to partition the domain using the partitioning line y = a into region I and region II.
The given integral will be sum of integral over region I and integral over region II.
Over region I : The limits of x is from x = 0 to  x = \sqrt {4ay} and limit of y is from y = 0 to y = a

Over region II : The limit of x is from x = 0 to x = 3ay and limit of y is from y = a to y = 3a

So, after changing the order, the integral converts to

 I = \int\limits_0^a {\int\limits_0^{\sqrt {4ay} } {f(x,y)dxdy} } + \int\limits_a^{3a} {\int\limits_0^{3a - y} {f(x,y)dxdy} }

Example 2 :

Change the order of double integration  \int\limits_0^1 {\int\limits_x^{\sqrt {2x - {x^2}} } {fdydx} }

Solution :

The region of integration is bounded by the curves

 C_1 : y = x

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

 C_3 : x = 0

 C_4 : x = 1

Let us plot all these curves
The region of integration is shown in figure

After changing the order, the given integral reduces to the form

 I = \int\limits_{y = c}^{y = d} {\int\limits_{x = {f_1}(y)}^{x = {f_2}(y)} {fdxdy} }

Let us draw the arrow parallel to x-axis and intersecting the region of integration arbitrarily.

The arrow first intersect the curve  C_2 at A. The equation of  C_2 has to be written in the form x = f(y)

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

 x^2 - 2x + y^2 = 0

 x = 1 \pm \sqrt {1 - {y^2}}


Since, A lies to left of x = 1. So,  x = 1 - \sqrt {1 - {y^2}}

The second curve it intersects is y = x which has to be written in the form  x = f_2(y) i.e. x = y

So, the limit of x is from  x = 1 - \sqrt {1 - {y^2}} to x = y

The arrow has to be moved from y = 0 to y = 1 to cover the whole domain

So, limit of y is from y = 0 to y = 1

So, after changing the order, the integral reduces to form

 I = \int\limits_0^1 {\int\limits_{1 - \sqrt {1 - {y^2}} }^y {fdxdy} }