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:
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 Ix:
`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 m2
b) Now to find the mass of the area, m.
m = kA, where A is the area.
`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:
` ~~ 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.
Please support IntMath!