Question

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.

Answer 1

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