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# Change of Order-2

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

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