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प्रश्न
Draw a rough sketch of the region {(x, y) : y2 ≤ 6ax and x 2 + y2 ≤ 16a2}. Also find the area of the region sketched using method of integration.
उत्तर
Given that: {(x, y) : y2 ≤ 6ax and x 2 + y2 ≤ 16a2}
Equation of Parabola is y2 = 6ax .....(i)
And equation of circle is x2 + y2 ≤ 16a2 .....(ii)
Solving equation (i) and (ii)
We get x2 + 6ax = 16a2
⇒ x2 + 6ax – 16a2 = 0
⇒ x2 + 8ax – 2ax – 16a2 = 0
⇒ x(x + 8a) – 2a(x + 8a) = 0
⇒ (x + 8a)(x – 2a) = 0
∴ x = 2a and x = – 8a. .....(Rejected as it is out of region)
Area of the required shaded region
= `2[int_0^(2"a") sqrt(6"a"x) "d"x + int_(2"a")^(4"a") sqrt(16"a"^2 - x^2) "d"x]`
= `2[sqrt(6"a") int_0^(2"a") sqrt(x) "d"x + int_(2"a")^(4"a") sqrt((4"a")^2 - x^2) "d"x]`
= `2sqrt(6"a") * 2/3 * [x^(3/2)]_0^(2"a") + 2[x/2 sqrt((4"a")^2 - x^2) + (16"a"^2)/2 sin^-1 x/(4"a")]_(2"a")^(4"a")`
= `(4sqrt(6))/3 * sqrt("a") [(2"a")^(3/2) - 0] + [xsqrt((4"a")^2 - x^2) + 16"a"^2 sin^-1 x/(4"a")]_(2"a")^(4"a")`
= `(8sqrt(12))/3 "a"^2 + [16"a"^2 8sin^-1 (1) - 2"a"sqrt(12"a"^2) - 16"a"^2 sin^-1 1/2]`
= `(16sqrt(13))/3 "a"^2 + [16"a"^2 * pi/2 - 2"a" * 2sqrt(3)"a" - 16"a"^2 * pi/6]`
= `(16sqrt(3))/3 "a"^2 + 8pi"a"^2 - 4sqrt(3)"a"^2 - 8/3 pi"a"^2`
= `((16sqrt(3))/3 - 4sqrt(3))"a"^2 + 16/3 pi"a"^2`
= `(4sqrt(3))/3 "a"^2 + 16/3 pi"a"^2`
= `4/3 (sqrt(3) + 4pi)"a"^2`
Hence, required area = `4/3(sqrt(3) + 4pi)"a"^2` sq.units
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