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The area of the plane region bJundedbythe cUlVes x+2y2 = 0 and x+3y2 =I is equal to ?

Latika Leekha
6 years ago
The given curves are x+2y2 = 0 and x+3y2 = I.
The first curve is x+2y2 = 0 which implies 2y2 = -x and hence y2 = -x/2, which is a parabola.
Second curve x+3y2 = I which is again a parabola as y2 = -1/3 (x-1).
Hence we are required to find the area between the two parabolas.
On solving the equations of the two parabolas, we get the points of intersection as (-2, 1) and (-2, -1).
Hence we set x = -2 and y = $\pm$ 1.
Hence the required area is | ∫ (-2y2 -1 + 3y2)dy | , where the integral runs from 0 to 1.
= | ∫ (y2 -1) dy| , where the integral runs from 0 to 1.
= | (1/3 -1) |
= 2/3.
Hence the area of the region bounded by the curves is 4/3.
Thnaks & Regrds
Latika Leekha