 ### IIT JAM PHYSICS

NEWTON LAWS OF MOTION
Tension
FRICTION
Friction
VELOCITY AND ACCELERATION
CENTRAL FORCES
Gravitation
UNIFORMLY ROTATING FRAME- CENTRIFUGAL AND CORIOLIS FORCES
CONSERVATION LAWS
Collision
CENTRE OF MASS AND VRIABLE MASS SYSTEMS
RIGID BODY DYNAMICS
FLUID DYNAMICS
COULOMB LAW AND ELECTRIC FIELD
GAUSS LAW OF ELECTROSTATICS AND APPLICATIONS
Capacitance
POLARIZATION OF DIELECTRICS
WORK AND ENERGY IN ELECTROSTATICS
BOUNDARY VALUE PROBLEMS
Capacitors
CURRENT ELECTRICITY
MAGNETOSTATICS
Ampere Law
MAGNETIC MATERIALS
DC CIRCUITS
RC Circuit
LR circuit
LC Circuit
AC CIRCUITS
AC Circuit
MAXWELL EQUATIONS and poynting vector
ELECTROMAGNETIC WAVES
REFLECTION AND REFRACTION OF EM WAVES AT THE INTERFACE OF TWO DIELECTRICS
Section 3: MATHEMATICAL PHYSICS
MULTIPLE INTEGRAL
VECTOR CALCULUS
DIFFERENTIAL EQUATIONS
MATRICES
Determinant
DIFFERENTIAL CALCULUS
Jacobian
FOURIER SERIES
PARTICLE NATURE OF WAVE
WAVE NATURE OF PARTICLE
H ATOM
POSTULATES OF QUANTUM MECHANICS
SCHRONDINGER WAVE WQUATION
NUCLEAR PHYSICS
SPECIAL THEORY OF RELATIVITY
SIMPLE HARMONIC OSCILLATION
DAMPED AND FORCED OSCILLATION
WAVES
Waves
GEOMETRICAL OPTICS
Thin Lens
INTERFERENCE
Thin Films
DIFFRACTION
Single Slit
Double Slit
POLARIZATION OF LIGHT
THERMAL EXPANSION
CALORIMETRY
Calorimetry
TRANSMISSION OF HEAT
1 of 2

# Change of Order-7

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