Question: Evaluate $\iint_R\frac{2xy^5}{\sqrt{1+x^2y^2-y^4}}dxdy$ where 'R' is the region of the triangle whose vertices are $(0,0),\, (1,1),\,(0,1)$ .
0

Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : Medium

am2(82) • 308 views
ADD COMMENTlink
 modified 11 months ago by awari.swati831 • 150 written 13 months ago by
0

$\text { Let O (0,0), A (1,1), B (0,1) be the vertices of the triangle OAB.} \\$

$\text { The equation of the line OA is} \\$

$\frac{x-0}{0-1} = \frac{y-0}{0-1} \text{ i.e. } x= y \\$

$\text {Now consider a strip parallel to x-axis } \\$

$\int_{y=0}^{1} \int_{x=0}^{y} \frac{2y^5x} {\sqrt {1- x^2y^2 - y^4} }dx dy \\$

$\int_{y=0}^{1} \int_{x=0}^{y} \frac{1}{y} \frac{2y^5*x} {\sqrt{\frac{1- y^4}{y^2} + x^2 } }dx dy \\$

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

$\text { Put } \frac{1- y^4}{y^2} + x^2 = t \\$

$2xdx = dt \\$

$= \int_{0}^{1} 2 y^4 \int \frac { dt} {2\sqrt t } \\$

$= \int_{0}^{1} y^4 (2\sqrt t ) dy \\$

$= 2 \int_{0}^{1} y ^4 \left [ \sqrt{\frac{1- y^4}{y^2} + x^2 }\right]_0^y dy \\$

$= 2 \int_{0}^{1} y ^4 \left [ \sqrt{\frac{1- y^4}{y^2}+ y^2} - \sqrt{\frac{1- y^4}{y^2} }\right] dy \\$

$= 2 \int_{0}^{1} y ^4 \left [ \frac{1}{y} - \sqrt{\frac{1- y^4}{y^2} }\right] dy \\$

$= 2 \int_{0}^{1} y ^3 - \sqrt{{1- y^4} } y^3 dy \\$

$= 2 \left [ \frac{y^4}{4} + \frac{1}{4} \frac{(1- y^4)^\frac{3}{2}}{\frac{3}{2}} )\right]_0^1 \\$

$= 2 [\frac{1}{4} - \frac{1}{4} \frac{2}{3} ] = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \\$

ADD COMMENTlink
 modified 11 months ago by awari.swati831 • 150 written 12 months ago by
Please log in to add an answer.