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.

First quadrant circle

In this example,

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

a) Finding Iy:

`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`


`= (2k)/15`

b) The mass of the area, m.

Now m = kA, where A is the area.


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



` = 2/3`

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

c) The radius of gyration:




` 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.