As usual, first we sketch the part of the curve in the first quadrant. It's a parabola, passing through (1, 1) and (0, 1). A "typical" rectangle is shown.

In this example,

`y_2= 1 − x^2`, and `y_1= 0`, `a = 0` and `b = 1`.

a) Finding *I*_{y}:

`I_y=k int_a^b x^2(y_2-y_1) dx`

`=k int_0^1x^2 [(1-x^2)-(0)] dx`

`=k int_0^1 [x^2-x^4] dx`

`=k[(x^3)/(3) - (x^5)/(5)]_0^1`

`=k(1/3-1/5)`

`= (2k)/15`

b) The **mass **of the area, *m*.

Now *m* = *kA*, where *A* is the area.

Now

`A=int_0^1 (1-x^2) dx`

`=[x-(x^3)/(3)]_0^1`

`=1-1/3`

` = 2/3`

So `m = kA = (2k)/3`

c) The **radius of gyration:**

`R_y=sqrt{(I_y)/(m)}`

`=sqrt{(2k//15)/(2k//3)}`

`=sqrt[1/5]`

` approx 0.447`

This means that if a mass of `(2k)/3` was placed `0.447` units from the `y`-axis, this would have the same **moment of inertia** as the original shape.