### 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-7

Example 1 :

Change of order of integration in double integral

$\int_{0}^{\pi/2} \int_{0}^{2a \cos \theta} f(r, \theta) dr d \theta$

Solution :

Let the compare the given integral to the integral in standard form

$\int_{0}^{\pi/2} \int_{0}^{2a \cos \theta} f(r, \theta) dr d \theta \equiv \int_{\theta = \alpha}^{\theta = \beta} \int_{r = f_{1}(\theta)}^{r = f_{2} (\theta)} f(r, \theta) dr d \theta$

The bounding curves are given by

$r = 0, \ r = 2a \cos \theta$

$\theta = 0, \ \theta = \pi/2$

Let us plot these curves to get the region of integration

After changing the order, the integral will be of the form

$I = \int_{r = r_{1}}^{r = r_{2}} \int_{\theta = g_{1}(r)}^{\theta = g_{2} (r)} f(r, \theta) d \theta dr$

To get the limit of $\theta,$ let us draw a circular arc with centre at the origin oriented in anticlockwise direction.

The arc intersects first $\theta = 0$ and then intersect $\theta = \cos^{-1} \dfrac{r}{2a}$. So, limit of $\theta$ is from $\theta = 0$ to $\theta = \cos^{-1} \dfrac{r}{2a}$. To find out limit of r, we must find out in what range the radius of arc should vary so that whole region of integration is transerved. r should vary from r = 0 to r = 2a.

So, after changing the order, the integral changes to

$I = \int_{0}^{2a} \int_{0}^{cos^{-1} \dfrac{r}{2a}} f(r, \theta) d \theta dr$

Example 2 :

Change the order of integration in the system of integrals

$\int_{0}^{\pi/2} \int_{a \cos \theta}^{a(1 + \cos \theta)} f(r, \theta) dr d\theta + \int_{\pi/2}^{\pi} \int_{0}^{a(1 + \cos \theta)} f(r, \theta)dr d \theta$

Solution :

Comparing the integral with its standard form

$\int_{\theta = \alpha}^{\theta = \beta} \int_{r = f_{1} (\theta)}^{r = f_{2}(\theta)} f(r, \theta) dr d \theta$

The region of integration is bounded by curves

$r = a \cos \theta \ \ \ \ \theta = 0$

$r = a(1 + \cos \theta), \ \ \ \ \theta = \pi$

After changing the order the integral is of the form

$I = \int_{r = r_{1}}^{r = r_{2}} \int_{\theta = f_{1} (r) }^{\theta = f_{2}(r)} f(r, \theta) d \theta dr$

To find the limit of $\theta$, let us draw an circular arc of radius r intersecting the region of integration arbitrary. The arc has two possibilities. If radius of arc is less than a it intersects circle and cardiod $(r = a(1 + \cos \theta))$ and if a < r < 2a, it intersects initial line and cardiod. So, we have to partition the domain of integration by a partitioning curve in the form of circular arc of radius a into region I and II.

For region, I let us draw a circular arc intersecting the region I arbitrary. The limit of $\theta = \cos^{-1} \dfrac{r}{a}$ to $\theta = \cos^{-1} \dfrac{r - a}{a}$ and r will vary from r = 0 to r = a.

For region II, let us draw a circular arc intersection the region II arbitrary. The limit of $\theta$ is from $\theta = 0$ to $\theta = \cos^{-1} \frac{r - a}{a}$ and r will vary from r = a to r = 2a.

So after changing the order, the integral will convert to

$I = \int_{0}^{a} \int_{\cos^{-1} \frac{r}{a}}^{\cos^{-1} \frac{r - a}{a}} f(r, \theta) d \theta dr + \int_{a}^{2a} \int_{0}^{\cos^{-1}\frac{r - a}{a}} f(r, \theta) d \theta dr$