Question: Evaluate the :
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$\int \int_R \frac{2x \ y^5}{\sqrt{1+ x^2\ y^2 – y^4}} \ dx \ dy$, where R is a triangle whose vertices are (0, 0), (1,1), (0,1).

Subject : Applied Mathematics 2

Topic : Double Integration

Difficulty: High

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modified 3 months ago  • written 3 months ago by gravatar for Ankit Pandey Ankit Pandey70
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Solution:

let $\quad I=\iint \frac{2 x y^{5}}{\sqrt{x^{2} y^{2}-y^{4}+1}} d x d y$

Region of integration: Triangle whose vertices are $(0,0),(1,1),(0,1)$

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The equation of lines from diagram are : $y=1, x=y$

$0 \leq x \leq y$

$0 \leq y \leq 1$

$\therefore I=\int_{0}^{1} \int_{0}^{y} \frac{2 x y^{5}}{\sqrt{x^{2} y^{2}-y^{4}+1}} d x\ d y$

$\therefore I=\int_{0}^{1} \int_{0}^{y} \frac{2 y^{5} x}{\sqrt{\left(1-y^{4}\right)+x^{2} y^{2}}} d x \ d y$

$\begin{aligned} &=\int_{0}^{1} \int_{0}^{y} \frac{2 y^{5} x}{\sqrt{\frac{1-y^{4}}{y^{2}+x^{2}}}} \cdot \frac{1}{y} d x \ d y \\ &=\int_{0}^{1} 2 y^{4}\left[\sqrt{\frac{1-y^{4}}{y^{2}}+x^{2}}\right]_{0}^{y} d y \\ &=\int_{0}^{1} 2 y^{4}\left[\frac{1}{y}-\frac{\sqrt{1-y^{4}}}{y}\right] d y \end{aligned}$

$\begin{aligned} &=2 \int_{0}^{1}\left[y^{3}-\sqrt{1-y^{4}} \cdot y^{3}\right] d y \\ &=2\left[\frac{y^{4}}{4}+\frac{1}{4} \cdot \frac{\left(1-y^{4}\right)^{3 / 2}}{3 / 2}\right]_{0}^{1} \\ &=2\left[\frac{1}{4}-\frac{1}{4} \cdot \frac{2}{3}\right] \\ \therefore I &=\frac{1}{6} \end{aligned}$

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written 3 months ago by gravatar for Ankit Pandey Ankit Pandey70
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