We sketch the parabola. The shaded area is the part that's rotating around the `x`-axis, and we have indicated a "typical" rectangle.

In this example, we are rotating the area around the `x`-axis. We need to express our function in terms of `y`.

Since we started with `y=x^2+1`, we solve for `x` and obtain:

`x=sqrt(y-1)`

We take the positive case only, as we are dealing with the first quadrant.

So the required values which we can use in the formula are:

`x_2= 2` (the "curve" furthest from the `y`-axis),

`x_1= sqrt(y-1)` (the curve closest to the `y`-axis), and

`c = 1` and `d = 5`; and `k=3`.

a) Finding *I*_{x}:

`I_x=k int_c^d y^2(x_2-x_1) dy`

`=3 int_1^5 y^2 [(2) - sqrt(y-1)] dy`

`=3 int_2^5 (2 y^2 - y^2 sqrt(y-1)) dy`

We use a computer algebra system to obtain:

`=45.5` kg m

^{2}

b) Now to find the **mass **of the area, *m*.

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

Now

`A=int_1^2 (x^2+1) dx - 1` (We need to subtract the area of the `1xx1=1` square below the shaded area.)

`=[(x^3)/(3)+x]_1^2 - 1`

`=(8/3+2)-(1/3+1) - 1`

` = 7/3`

So `m = kA = 3xx7/3 = 7` kg

c) The **radius of gyration:**

`R_x=sqrt{(I_x)/(m)}`

`=sqrt{(45.5)/(7)}`

` ~~ 2.6` m

This means that if a single mass of `7` kg was placed `2.6` m from the `x`-axis, this would have the same **moment of inertia** as the original shape when rotating around the `x`-axis.

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