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## Solids of Revolution: DisksThe 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 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.http://www.physics.brocku.ca/ http://atarnotes.com/forum http://www.nabla.hr (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. |