Question

Find the area of the region bounded by `x^2 = 4y, y = 2, y = 4`, and the `y`-axis in the first quadrant.

Options

• `(8sqrt(2))/3` sq.units

• `32/2` sq.units

• `((32 - 8sqrt(2))/3)` sq.units

• `((32 + 8sqrt(2))/3)` sq.units

Answer 1

`((32 - 8sqrt(2))/3)` sq.units

Explanation:

The given curve `x^2 = 4x` is a parabola with vertex at (0, 0) Also since it contains only even coures of x, it is symmetrical about `y-axis.

`y` = 2 and `y` = 4 are straight lines parallel to `x`-axis at a positive distance of 2 and 4 from it respectively.

∴ Required area = area ABCD

= `int_2^4 xdy = int_2^4 2sqrt(y)dy = int_2^4 sqrt(y) dy`

= `2[y^(3/2)/(3/2)]_2^4 = 2 xx 2/3[4^(3/2) - 2^(3/2)] = 4/3 (8 - 2sqrt(2))`

= `((32 - 8sqrt(2))/3)` sq.units.