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Change of Order-4

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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} }