The hard part about getting volumes from integration is figuring
out the cross-sectional area $A(x)$. In general, this
requires some knowledge of geometry. But if we have a solid of revolution, the area of
the cross-section is easy to determine - the cross-section is a
circle. The cross-sectional solids are either disks (solid
circles, sliced thinly) or washers (circles with the middle
missing, sliced thinly). First we will discuss the disks.

Solids of revolution are
obtained by taking a curve, say $f(x)$, and rotating the curve
around some axis, called the axis of
revolution. The first two examples below show
$f(x)$ rotated around the $x$-axis, while the third shows $f(x)=\sin
x$ between $0$ and $\tfrac \pi 2$ rotated around the $y$-axis.
Curves may also be rotated around other lines.

We get cross-sectional disks, with area $A=\pi r^2$, where the
radius $r$ is $f(x)$ or $g(y)$. (Don't
go further until you see why the curve is the radius.)
When rotating around the $x$-axis, for any value of $x$ the area of
the circle is $A(x) = \pi \left(f(x)\right)^2$ and its thickness is
$\Delta x$. Thus the volume of each slice is approximately $
\pi \left(f(x)\right)^2\,\Delta x$. Similarly, when rotating
around the $y$-axis, the volume of each slice is approximately $ \pi
\left(g(y)\right)^2\,\Delta y$. After we add up the slices and
take a limit we get the total volume as follows, depending upon
whether we are rotating the curve around the $x$-axis or the
$y$-axis.

There is no need to
memorize this integral; the area of the cross-section is
your integrand.

The volume in the third example above would be
$$\pi\int_0^{\tfrac \pi 2}\left(\sin^{-1}(y)\right)^2\,dy,$$
since the curve $y=f(x)=\sin x$ is equivalent to
$x=g(y)=\sin^{-1}y$, and we need our function to be in terms of
$y$ when rotating around the $y$-axis. Fortunately, you do
not need to evaluate this integral.