More Applications of Integrals

(You will see problems like this and work through them in the second semester of calculus.)

Volumes

In many cases, we can get volumes by integration, too. Suppose we have a solid, like a sphere or a cone. To figure out its volume, we put it through a meat slicer, figure out the (approximate) volume of each slice, and then add up the slices. If each slice has area $A(x)$ and thickness $\Delta x$, then each slice has volume $A(x)\Delta x$. Add them up and take a limit to get $$\hbox{Volume} = \int_a^b A(x)\, dx = \lim_{n \to \infty} \, \sum_{i=1}^n\, A(x_i^*)\, \Delta x,$$ where the left-most slice is at $x=a$, the right-most slice is at $x=b$, and the cross-sectional area at position $x$ is given by the function $A(x)$.

As explained in the video, if we apply this method to a cone of height 1 whose base is a circle of radius 1, we get the integral $\pi \int_0^1 x^2 dx$.

Moment of Inertia

The moment of inertia of a particle is an indicator of how much torque you need to rotate it around the origin. For a point particle of mass $m$ a distance $r$ from the origin, the moment of inertia is $mr^2$. But what value of $r$ do we use for a big object like a bar? The answer is to:

1. Break the bar into $n$ tiny pieces, so that the distance from the origin is almost constant in each piece.
2. Figure out the mass of each piece,
3. Figure out how far each piece is from the origin,
4. Approximate the moment of inertia of each piece,
5. Add up the moments of inertia of all the pieces, and
6. Take a limit to get the exact answer. As with areas, distances and volumes, this limit is an integral.

For a uniform thin bar of mass 1kg and length 1m, we find that each piece has mass $1 \Delta x$ and distance $x_i^*$ from the origin, so the total moment of inertia is $$\lim_{n \to \infty} \,\sum_{i=1}^n\, (x_i^*)^2\, \Delta x = \int_0^1 x^2\, dx.$$ This is the same integral that gives the area under a parabola, or the volume of a cone.