DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA

Change of Order-3

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Example 1 :

Change the order of double integration  \int\limits_{ - 1/\sqrt 2 }^{1/\sqrt 2 } {\int\limits_{ - \sqrt {1 - 2{x^2}} }^{\sqrt {1 - 2{x^2}} } {fdy\,dx} }

Solution :

The given integral is

 I = \int\limits_{ - 1/\sqrt 2 }^{1/\sqrt 2 } {\int\limits_{ - \sqrt {1 - 2{x^2}} }^{\sqrt {1 - 2{x^2}} } {fdy\,dx} }

Comparing the two integral, the curves bounding the region of integration are given by

 {C_1}:y = - \sqrt {1 - 2{x^2}}

 {C_2}:y = \sqrt {1 - 2{x^2}}

 {C_3}: x  = - \dfrac{1}{{\sqrt 2 }}

 {C_4}: x = \dfrac{1}{{\sqrt 2 }}

 C_1 &  C_2 are the part of ellipse


 2x^2 + y^2 = 1


The region of integration is shown in figure

After changing the order, the given integral converts to

 I = \int\limits_{y = c}^{y = d} {\int\limits_{x = {g_1}(y)}^{x = {g_2}(y)} {fdx\,dy} }

To get the limit of x, let us draw an arrow parallel to x-axis intersecting the region of integration arbitrarily.

The arrow enters the region at A and leaves at point B.

Let us write the equation of curve at A and B in the form x = g(y)

At  A, x = - \sqrt {\dfrac{{1 - {y^2}}}{2}} and at  B, x = \sqrt {\dfrac{{1 - {y^2}}}{2}}


The arrow has to be moved from y = –1 to y = 1 so as the traverse the whole region of integration.

Hence, the unit of x is from  - \sqrt {\dfrac{{1 - {y^2}}}{2}} to   \sqrt {\dfrac{{1 - {y^2}}}{2}} and that of y from –1 to +1.

So, after changing the order

 I = \int_{ - 1}^1 \int_{ \sqrt{\dfrac{1-y^2}{2}}}^{\sqrt{\dfrac{1-y^2}{2}}} fdxdy

Example 2 :

Change the order of double integration.  I = \int\limits_1^2 {\left( {\int\limits_x^{2x} {fdy} } \right)dx}

Solution :

The given integral is

 I = \int\limits_1^2 {\left( {\int\limits_x^{2x} {fdy} } \right)dx}

 \int\limits_{x = a}^{x = b} {\left( {\int\limits_{y = {f_1}(x)}^{y = {f_2}(x)} {fdy} } \right)dx}

Comparing the two integral, the curves bounding the region of integration are given by

 C_1 : y = x

 C_2 : y = 2x

 C_3 : x = 1

 C_4 : x = 2

Let us draw these curves to get the region of integration.
After changing the order, the given integral converts to

 I = \int\limits_{y = c}^{y=d}{\int\limits_{x = g(y)}^{x = {g_2}(y)}{fdxdy}}


To get the limit of x, let us draw an arrow parallel to x-axis intersecting the region of integration arbitrarily.
From the region of integration plotted in figure we find that if arrow is drawn below line, y = 2, the curves intersected are x = 1 and y = x but if arrow is drawn above line y = 2 the curves intersected are y = 2x and x = 2.
In this case the region of integration has to be partitioned into region I and region II by partition line y = 2.
So, the given integral will be broken into two integrals corresponding to I and II while changing the order. For region I, the limit of x varies from x = 1 to x = y and that of y varies from y = 1 to y = 2.
For region II, the limit of x varies from  x = \dfrac{y}{2} to x = 2 and that of y varies from y = 2 to y = 4.
So, after changing the order, the integral changes to

 I = \int\limits_1^2 {\int\limits_1^y {fdxdy} } + \int\limits_2^4 {\int\limits_{y/2}^2 {fdxdy} }