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The area of the circle x2+y2=16, exterior to the parabola y=6x -

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Question

The area of the circle `x^2 + y^2 = 16`, exterior to the parabola `y = 6x`

Options

  • `4/3 (4pi - sqrt(3))`

  • `4/3 (4pi + sqrt(3))`

  • `4/3 (8pi - sqrt(3))`

  • `4/3 (8pi + sqrt(3))`

MCQ

Solution

`4/3 (8pi - sqrt(3))`

Explanation:

The given curve are

`x^2 + y^2` = 16  ......(1)

`y^2 = 6x`  ......(2)

Putting `y^2 = 6x` in (1)

`x^2 + 6x` = 16 or `x^2 + 6x - 16` = 0

`(x + 8)(x - 2)` = 0

∴ `x = - 8, 2`

But `x ≠ - 8`

∴ `x` = 0

From (2), `y^2 = 6x, y = +-`

Area of the whole circle

= `4 int_0^1 sqrt(16 - x^2)  dx`

= `4[x/2 sqrt(16 - x^2) + 16/2 sin^-1 x/4]_0^4`

= `4[0+ 8(sin^-1  y/4 - 0)] = 32 sin^-1 1`

= `32 xx pi/2 = 16 pi` sq.units 

Now circle and parabola intersects at `P(2, 2sqrt(3))` and `theta(2, -2sqrt(3))`

Smaller area enclosed by circle and parabola,

Area of region OθAP = 2 × Area of region OMAP

= 2[Area of region OMP + Area of region MAP]

= `2int_0^2 y_1  dx + int_2^4 y_2  dx`

`y = 6x` or `y = sqrt(6x)`

Then `y`, is for parabola, `y = sqrt(16 - x^2)`

`y` is for circle `x^2 + y^2` = 16, `y = sqrt(16 - x^2)`

= `2[int_0^2 sqrt(6x)dx + int_2^4 sqrt(16 - x^2)  dx]`

= `2[sqrt(6) * 2/3 [x^(3/2)]_0^2 + [x/2 sqrt(16 - x^2) + 16/2  sin^-1  x/4]_2^1]`

= `2[2/3 sqrt(6) * (2^(3/2) - 0) + (10 + 8sin^-1  1) - (sqrt(12) + 8sin^-1  1/2)]`

= `4/3 sqrt(6) * sqrt(8) + 2(8 xx pi/2 - sqrt(12) - 8 * pi/2)`

= `16/3 sqrt(3) + 8pi - 2sqrt(12)`

= `8/3 pi = ((16sqrt(3))/3 - 4sqrt(3)) + 16/3 pi`

= `(4sqrt(3))/3 + 16/3  pi`

Common area exprior to the parabola `y^2 = 6x`

= `16pi - ((4sqrt(3))/3 + 16/3  pi)`

= `32/2 pi - (4sqrt(3))/3`

= `4/3 (8pi - sqrt(3))`

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