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Question
Find the area of the region bounded by the curve y2 = 2x and x2 + y2 = 4x.
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Solution
Equations of the curves are given by x2 + y2 = 4x ......(i)
And y2 = 2x ......(ii)
⇒ x2 – 4x + y2 = 0
⇒ x2 – 4x + 4 – 4 + y2 = 0
⇒ (x – 2)2 + y2 = 4
Clearly it is the equation of a circle having its centre (2, 0) and radius 2.
Solving x2 + y2 = 4x and y2 = 2x
x2 + 2x = 4x
⇒ x2 + 2x – 4x = 0
⇒ x2 – 2x = 0
⇒ x(x – 2) = 0
∴ x = 0, 2
Area of the required region
= `2[int_0^2 sqrt(4 - (x - 2)^2) "d"x - int_0^2 sqrt(2x) "d"x]`
∴ Parabola and circle both are symmetrical about x-axis.
= `2[(x - 2)/2 sqrt(4 - (x - 2)^2) + 4/2 sin^-1 (x - 2)/2]_0^2 - 2*sqrt(2)* 2/3 [x^(3/2)]_0^2`
= `2[(0 + 0) - (0 + 2 sin^-1 (-1)] - (4sqrt(2))/3 [2^(3/2) - 0]`
= `-2 xx 2 * (- pi/2) - (4sqrt(2))/3 * 2sqrt(2)`
= `2pi - 16/3`
= `2(pi - 8/3)` sq.units
Hence, the required area = `2(pi - 8/3)` sq.units
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