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:

Break the bar into $n$ tiny pieces, so that the distance from the
origin is almost constant in each piece.

Figure out the mass of each piece,

Figure out how far each piece is from the origin,

Approximate the moment of inertia of each piece,

Add up the moments of inertia of all the pieces, and

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.