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Change of Order-4

MATHEMATICAL PHYSICS FOR IIT JAM Change of Order Change of Order-4

Example 1 :

Change the order of integration. $I = \int\limits_0^{2a} {\int\limits_{\sqrt {2ax - {x^2}} }^{\sqrt {2ax} } {fd} ydx}$

Solution :

The given integral is

$I = \int\limits_0^{2a} {\int\limits_{\sqrt {2ax - {x^2}} }^{\sqrt {2ax} } {fd} ydx}$

$I = \int\limits_{x = a}^{x = b} {\int\limits_{y = {g_1}(x)}^{y = {g_2}(x)} {fdydx}}$

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

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

${C_2}:y = \sqrt {2ax}$

${C_3}:x = 0$

${C_4}:x = 2a$

Let us trace all the curves to get region of integration

After changing the order of integration, the integral changes to

$I =\int\limits_{y = c}^{y=d} {\int\limits_{x={g_1}(y)}^{x = {g_2}(y)}{fd} xdy}$

The domain has to be partitioned by line y = a into three regions as shown in figure.

For region I, x varies from $x = \dfrac{{{y^2}}}{{2a}}$ to $x = a - \sqrt {{a^2} - {y^2}}$ and y varies from y = 0 to y = a

For region II, x varies from $x = a + \sqrt {{a^2} - {y^2}}$ to x = 2a and y varies from y = 0 to y = a

For region III, x varies from $x = \dfrac{{{y^2}}}{{2a}}$ to x = 2a and y varies from y = a to y = 2a

So, after changing the order, the integral becomes

$I = \int\limits_0^a {\int\limits_{{y^2}/2a}^{a - \sqrt {{a^2} - {y^2}} } {fdxdy} } + \int\limits_0^a {\int\limits_{a + \sqrt {{a^2} - {y^2}} }^{2a} {fdxdy} } + \int\limits_a^{2a} {\int\limits_{{y^2}/2a}^{2a} {fdxdy} }$