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-1

Lesson Progress
0% Complete
video

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} }