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Question: Change the order of the integration in the following $\int_{0}^{1/2}\int_{0}^{\sqrt{1-4y^2}}\frac{(1+x^2)}{\sqrt{1-x^2}\,\sqrt{1-x^2-y^2}}dx\,dy$ .
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Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : High

am2(82) • 241 views
 modified 6 months ago by awari.swati831 • 140 written 8 months ago by
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$\text { The region of integration is bounded by x = 0 i.e. the y-axis } \\$

$x = \sqrt{1-4y^2} \Longrightarrow x^2 = 1- 4y^2 \\$

$x^2 + \frac{y^2} {\frac{1}{4}} = 1 \\$

$\text { It is an ellipse with semi major axis 1 and semi minor axis } \frac{1}{2} \text {y = 0 and y = } \frac{1}{2} \text { Thus the region of integration is the first quadrant of the above ellipse. } \\$

$\text { To change the order of integration, consider a strip parallel to y- axis. On this strip y varies from y = 0 to y =\frac{\sqrt{1-x^2}} {2} and x varies from 0 t0 1. } \\$

$I = \int_{x=0}^{1} \int_{y=0}^{\sqrt{\frac{1+x^2}{4}}} \frac {(1+x^2)} {\sqrt{1-x^2}} \frac{dy dx} {\sqrt{1- x^2 - y^2} } \\$

$I = \int_{x=0}^{1} \frac {(1+x^2)} {\sqrt{1-x^2}} \left [ sin^{-1} \frac{y} {\sqrt{1- x^2} } \right]_0^{\sqrt{\frac{1+x^2}{4}}}dx \\$

$I = \int_{x=0}^{1} \frac {(1+x^2)} {\sqrt{1-x^2}} \left [ sin^{-1} \frac{1} {2} - sin^{-1} 0 \right]dx \\$

$= \frac{\pi}{6} \int_{x=0}^{1} \frac {(1+x^2)} {\sqrt{1-x^2}} dx \\$

$\text { Put x = } sin\theta, dx = cos\theta d\theta \\$

$I = \frac{\pi}{6} \int_{0}^{\frac{\pi}{2}} \frac {(1+sin^2\theta)} {\sqrt{1-sin^2\theta}}cos\theta d\theta \\$

$I = \frac{\pi}{6} \int_{0}^{\frac{\pi}{2}} \frac {(1+sin^2\theta)} {cos\theta}cos\theta d\theta \\$

$I = \frac{\pi}{6} [ \int_{0}^{\frac{\pi}{2}} d\theta + \int_{0}^{\frac{\pi}{2}} \frac {(1-cos2\theta)} {2}d\theta ] \\$

$I = \frac{\pi}{6} \left[ \theta +\frac{1}{2} (\theta -\frac {sin2\theta} {2}\right]_0^\frac{\pi}{2} \\$

$I = \frac{\pi}{6} \left[ \frac{\pi}{2} + \frac{1}{2}\frac{\pi}{2}\right] \\$

$I = \frac{\pi}{6} \frac{3\pi}{4} \\$

$= \frac{(\pi)^2}{8} \\$