Question: Evaluate $\iint_R y\,dxdy$ where 'R' is the region bounded by $y^2=4x$ & $x^2=4y$
0

Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : Medium

am2(82) • 377 views
 modified 11 months ago by awari.swati831 • 150 written 13 months ago by
0

$\text { The two parabola's intersect at O(0,0) and A (4,4).} \\$

$\text {Point of intersection : } \\$

$y^2 = 4x , x^2 = 4y \\$

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

$x^2 = 4y \\$

$\frac { y^4} {16} = 4y \\$

$y^4 = 64 y \\$

$y(y^3- 64) = 0 \\$

$y= 0 , y^3-64 = 0 \text { i.e. } y= 4\\$

$\therefore y = 4, x = 4 \\$

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

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

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

$= \int_{y=0}^{4} \left [ \sqrt{4} y^\frac{3}{2} - \frac{y^3}{4}\right ]dy \\$

$= \left [ \sqrt{4} \frac{y^\frac{5}{2}}{5/{2}} - \frac{y^4}{16}\right ]_0^4 \\$ ) $= \frac{2\sqrt4}{5} 4^{\frac{5}{2}} - \frac{(4)^4}{16} \\$

$= \frac{48}{5} \\$