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Exercise: Area Between Two Curves · Q22

Q.Find the area of the smaller region bounded by the ellipse x29+y24=1\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 and the line x3+y2=1\dfrac{x}{3} + \dfrac{y}{2} = 1.

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Both the ellipse and the line pass through (3,0)(3,0) and (0,2)(0,2); the ellipse's arc bulges above the chord between them.

The ellipse x29+y24=1\dfrac{x^2}{9}+\dfrac{y^2}{4}=1 has a=3a=3, b=2b=2, so its upper arc is y=239−x2y=\dfrac23\sqrt{9-x^2}. The line x3+y2=1\dfrac{x}{3}+\dfrac{y}{2}=1 can be written as y=2−2x3y=2-\dfrac{2x}{3}. Both curves pass through (3,0)(3,0) and (0,2)(0,2) -- these are the two intersection points -- and for 0<x<30<x<3 the ellipse's arc lies above the straight chord (a straight line joining two points of a convex arc always lies inside the arc). The enclosed area is

Area=∫03[239−x2−(2−2x3)]dx=23∫039−x2 dx−∫03(2−2x3)dx.\text{Area} = \int_0^3\left[\frac23\sqrt{9-x^2}-\left(2-\frac{2x}{3}\right)\right]dx = \frac23\int_0^3\sqrt{9-x^2}\,dx - \int_0^3\left(2-\frac{2x}{3}\right)dx. …

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