# Multiple integration to find the moment of inertia of a cylinder.

I was assigned a question asking: Find the moment of inertia of a circular cylinder with base $a$ and height $h$ about the diameter of the base. We are working through applications of multiple integrals currently so I am trying to solve this using cylindrical coordinates.

In my book it gives us the idea that Moment of Inertia $= \int \int \int D^2\cdot\text{density}\ dV$.

With $D$ being the perpendicular distance from the volume element $dV$ to the axis of rotation L.

I am confused at how to find what $D$ is in this question.

Can anyone help?

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You are rotating the cylinder about an axis perpendicular to the axis of symmetry. In this case, polars may not be your best bet.

Align the $x$ axis with the axis of rotation. For each point on the axis of rotation, which extands between $x \in [-a,a]$, there is a corresponding rectangular cross-section of the cylinder. Each point of the cross section is at a distance $D$ from that point of the axis of rotation. That rectangle has dimensions $2 \sqrt{a^2-x^2} \times h$. A point within that rectangle has distance from that point on the axis $D$ equal to $\sqrt{y^2+z^2}$. The get the element of inertia from this rectangle, integrate $y^2+z^2$ over the rectangular area. Then integrate the collection of elements of inertia from rectangles like this over $x$.

Setting this up: the element of inertia $dI(x)$ is an integral over that rectangular area:

$$dI(x) = \rho dx \int_0^h dz \: \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} dy \: (y^2+z^2) = \rho dx \frac{2 h}{3} \sqrt{a^2-x^2} (a^2+h^2-x^2)$$

Once you determine this, integrate over $x$ to get the total moment of inertia:

$$I = \rho \frac{2 h}{3} \int_{-a}^a dx \: \sqrt{a^2-x^2} (a^2+h^2-x^2) = \frac{1}{12} M (3 a^2+4 h^2)$$

I leave the details of evaluating the integrals to the reader.

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