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

Topic : Double integration

Difficulty : Medium

am2(82) • 221 views
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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 8 months ago by gravatar for smitapn612 smitapn6120
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$ \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} \\ $

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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 7 months ago by gravatar for smitapn612 smitapn6120
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