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

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

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