+397 Hi OldTimer here's one method.
Sketch the graph and take a strip parallel with the x - axis, distance y from the axis and width delta y, from the curve on the left to the vertical at x = a on the right.
Spin this around the x - axis, forming a cylinder.
Vol of cylinder =
\(\displaystyle \{ \pi(y+\delta y )^{2}-\pi y^{2}\}(a-x)=2\pi y\delta y(a-x),\)
(ignoring 2nd order terms, here and elsewhere).
All points on the cylinder are distance y from the x - axis so its moment of inertia (or second moment) about the x - axis
\(\displaystyle = 2\pi y \delta y(a-x).y^{2}=2\pi y^{3}(a-x)\delta y.\)
This now has to be summed for all cylinders between y = 0 and y = a and taken in the limit as delta y tends to zero.
That gets you the integral
\(\displaystyle \int^{a}_02\pi y^{3}(a-\frac{y^{2}}{a})dy=\frac{\pi a^{5}}{6}.\)
Now express this in the form Mk^2, where M is the mass of the solid and k is its radius of gyration.
If unit mass is equal to unit volume, then M is simply the volume of the solid. So,
\(\displaystyle M = \pi\int^{a}_{0}y^{2}dx = \pi \int^{a}_{0}ax dx = \frac{\pi a^{3}}{2}.\)
Finally then, the moment of inertia of the solid = \(\displaystyle \frac{\pi a^{5}}{6}=\frac{\pi a^{3}}{2}.\frac{a^{2}}{3}=M\left(\frac{a^{2}}{3}\right),\)
meaning that the square of the radius of gyration is (a^2)/3.
An alternative method would be to split the solid into circular discs perpendicular to the x - axis, and to sum these from x = 0 to x = a.
If you do that, you could make use of the standard result for the moment of inertia of a disc.
+397 Hi OldTimer here's one method.
Sketch the graph and take a strip parallel with the x - axis, distance y from the axis and width delta y, from the curve on the left to the vertical at x = a on the right.
Spin this around the x - axis, forming a cylinder.
Vol of cylinder =
\(\displaystyle \{ \pi(y+\delta y )^{2}-\pi y^{2}\}(a-x)=2\pi y\delta y(a-x),\)
(ignoring 2nd order terms, here and elsewhere).
All points on the cylinder are distance y from the x - axis so its moment of inertia (or second moment) about the x - axis
\(\displaystyle = 2\pi y \delta y(a-x).y^{2}=2\pi y^{3}(a-x)\delta y.\)
This now has to be summed for all cylinders between y = 0 and y = a and taken in the limit as delta y tends to zero.
That gets you the integral
\(\displaystyle \int^{a}_02\pi y^{3}(a-\frac{y^{2}}{a})dy=\frac{\pi a^{5}}{6}.\)
Now express this in the form Mk^2, where M is the mass of the solid and k is its radius of gyration.
If unit mass is equal to unit volume, then M is simply the volume of the solid. So,
\(\displaystyle M = \pi\int^{a}_{0}y^{2}dx = \pi \int^{a}_{0}ax dx = \frac{\pi a^{3}}{2}.\)
Finally then, the moment of inertia of the solid = \(\displaystyle \frac{\pi a^{5}}{6}=\frac{\pi a^{3}}{2}.\frac{a^{2}}{3}=M\left(\frac{a^{2}}{3}\right),\)
meaning that the square of the radius of gyration is (a^2)/3.
An alternative method would be to split the solid into circular discs perpendicular to the x - axis, and to sum these from x = 0 to x = a.
If you do that, you could make use of the standard result for the moment of inertia of a disc.
