Question

Find the area of the region lying in the first quadrant and enclosed by the x-axis, the line y = x and the circle x² + y² = 32.

Answer 1

We have y = 0, y = x and the circle x² + y² = 32 in the first quadrant.

Solving y = x with the circle

x² + x² = 32

x² = 16

x = 4   ...(In the first quadrant)

When x = 4, y = 4 for the point of intersection of the circle with the x-axis.

Put y = 0

x² + 0 = 32

`x = +- 4sqrt(2)`

So, the circle intersects the x-axis at `(+- 4sqrt(2), 0)`.

From the above figure, area of the shaded region,

`A = int_0^4 xdx + int_4^(4sqrt(2)) sqrt((4sqrt(2))^2 - x^2) dx`

= `[x^2/2]_0^4 + [x/2 sqrt((4sqrt(2))^2 - x^2) + (4sqrt(2))^2/2 sin^-1  x/(4sqrt(2))]_4^(4sqrt(2))`

= `[16/2] + [0 + 16 sin^-1 1 - 4/2 sqrt((4sqrt(2))^2 - 16^2) - 16 sin^-1  4/(4sqrt(2))]`

= `8 + [16 π/2 - 2sqrt(16) - 16  π/4]`

= 8 + [8π – 8 – 4π]

= 4π sq. units