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Question

Find the area bounded by the curves x = y2 and x = 3 − 2y2.

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Solution




x = y2 is a parabola opening towards positive x-axis , having vertex at O (0,0) and symmetrical about x-axis
x = 3 − 2y2 is a parabola opening negative x-axis, having vertex at A (3, 0) and symmetrical about x-axis, cutting y-axis at B and B'

Solving the two equations for the point of intersection of two parabolas

x =y2x=3-2y2y2 =3-2y2 3y2 =3y =±1y=1 ,x=1 and y=-1 x=1E1,1 and F1,-1are two points of intersection. The curve character changes at E and F .Draw EF parallel to y-axis.C(1, 0) is the point of intersection of EF ith x-axisSince both curves are symmetrical about x-axis ,Area of shaded region OEAFO = 2 Area OEAO=2Area OECO +area CEAC .....1Area OECO =o1y1 dx where Px, y1 is a point on x=y2 =o1y1 dx as y1 >0=o1x dx=x323201 =23 sq units .....2area CEAC =13y2 dx where Qx, y2 is a point on x=3-2y2 =13y2 dx as y2 >0 =133-x2dx=12133-x dx=12-3-x323213=12×230+232=2×222×3=43 sq. units .....3 From 1, 2 and 3, we get Therefore, area of Shaded region OEAFO =223 +43 =2×2 =4 sq units

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