### 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
Capacitors
BOUNDARY VALUE PROBLEMS
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 EQUATION
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-6

Example 1 :

Evaluate $\int \int \cos (x + y) dx dy$ over the region enclosed by $x = 0, y = \pi$ and $y = x$.

Solution :

The region of integration is shown in the figure

$I = \int \int \cos (x + y)dx dy$

$= \int_{0}^{\pi} \int_{0}^{y} \cos(x + y) dx dy$

$= \int_{0}^{\pi} \left [ \sin (x + y) \right ]_{0}^{y} dy$

$= \int_{0}^{\pi} \left ( \sin 2y - \sin y \right ) dy$

$= - \dfrac{1}{2} \left . \cos 2y + \cos y \right |_{0}^{\pi}$

$= -2$

Example 2 :

Evaluate the integral $\int \int xe^{x^{2} - y^{2}} dx dy$ over the region bounded by the lines y = x, y = x – 1, y = 0 and y = 1.

Solution :

The region of integration is shown in figure. It is easier to evaluate the integral first w.r.t. x and then w.r.t. y.

$I = \int_{0}^{1} \int_{y}^{y + 1} xe^{x^{2} - y^{2}} dx dy$

$= \dfrac{1}{2} \int_{0}^{1} e^{-y^{2}} \left [ e^{x^{2}} \right ]_{y}^{y + 1} dy$

$= \dfrac{1}{2} \int_{0}^{1} \left (e^{2y + 1} - 1 \right ) dy$

$= \dfrac{1}{2} \left [ \dfrac{e}{2} e^{2y} - y \right ]_{0}^{1}$

$= \dfrac{1}{4} [e^{3} - e - 2]$

Example 3 :

Evaluate $\int \int (x^{2} + y^{2}) dx dy$ over the region bounded by xy = 1, y = 0, y = x and x = 2.

Solution :

The region of integration is as shown in figure. It has to be partitioned by line x = 1, before integration first w.r.t. y and then w.r.t x,

$I = \int \int (x^{2} + y^{2})dy dx$

$= \int \int (x^{2} + y^{2}) dy dx + \int \int (x^{2} + y^{2})dy dx = I_{1} + I_{2}$

$I_{1} = \int_{0}^{1} \int_{0}^{x} (x^{2} + y^{2})dy dx$

$= \int_{0}^{1} \left [ x^{2} y + \dfrac{y^{3}}{3} \right ]_{0}^{x} dx$

$= \int_{0}^{1} \dfrac{4}{3} x^{3} dx$

$= \dfrac{4}{3} \left [ \dfrac{x^{4}}{4} \right ]_{0}^{1} = \dfrac{1}{3}$
$I_{2} = \int_{1}^{2} \int_{0}^{1/x} (x^{2} + y^{2}) dy dx$

$= \int_{1}^{2} x^{2} y + \left . \dfrac{y^{3}}{3} \right |_{0}^{1/x} dx$

$= \int_{1}^{2} \left ( x + \dfrac{1}{3x^{3}} \right ) dx$

$= \left [ \dfrac{x^{2}}{2} - \dfrac{1}{6x^{{2}}} \right ]_{1}^{2}$

$= \dfrac{39}{24}$

So, $I = I_{1} + I_{2}$

$= \dfrac{47}{24}$