We have seen how to calculate the area below graphs and between two graphs by means of integration. Now we will take a look on how to calculate the volume of a solid of revolution by means of integration. A solid of revolution is obtained by revolving the graph around an axis over #360^{\circ}# (or #2 \pi# rad).
The surface #S# is enclosed by the function #\blue f#, the #x#-axis and the lines #x=a# and #x=b#.
The volume of the body #\orange B#, resulting from #S# revolving around the #x#-axis is equal to
\[V(\orange B)=\pi \cdot \int_a^b (\blue{f(x)})^2 \; \dd x\]
We prove this by showing that the derivative of the volume is equal to #\pi (\blue{f(x)})^2#, since then we have that #V(x)# is an anti-derivative of #\pi (\blue{f(x)})^2#.
For #V'(x)# we have, according to the definition of the derivative,
\[V'(x) = \frac{V(x+h)-V(x)}{h} \text{ with } h \to 0\]
For small #h# we can approach #V(x+h)-V(x)# by a cylinder with ground surface with area #\pi (\blue{f(x)})^2# and height #h#. Hence
\[V(x+h)-V(x)= \pi (\blue{f(x)})^2 \cdot h\]
This means that
\[\frac{V(x+h)-V(x)}{h}=\pi (\blue{f(x)})^2\]
Hence we have what we wanted to prove #V'(x)=\pi (\blue{f(x)})^2#. Therefore we have #V(x)=\pi \int (\blue{f(x)})^2 \; \dd x#.
Above we have shown how to calculate the volume of a solid of revolution. It's also possible to calculate the surface of a solid of revolution. If you want to calculate the surface of the solid #\orange{B}# as described above, it will be equal to \[A(\orange{B})=2\pi\cdot \int_a^b \blue{f}(x)\cdot \sqrt{1+(\purple{f'}(x))^2}\]
If we analyze the function #f(x)=\frac{1}{x}# on the interval #\left[0,\infty\right)#, we notice something interesting. The volume of the solid of revolution, which looks like a horn, is equal to
\[\pi \cdot \int_1^{\infty}\left(\frac{1}{x}\right)^2\, \dd x=\pi\cdot \left[-\frac{1}{x}\right]_{1}^{\infty}=\pi\cdot \left(-\frac{1}{\infty}-(-1)\right)=\pi,\] we note that we make use of #\frac{1}{\infty}=0#.
If we calculate the surface of the solid of revolution, we get
\[2\pi \int_1^{\infty}\frac{1}{x}\cdot \sqrt{1+\left(-\frac{1}{x^2}\right)^2}\, \dd x\]
Since #\left(-\frac{1}{x^2}\right)^2# is always positive #\sqrt{1+\left(-\frac{1}{x^2}\right)^2}# is alway strictly bigger than #1#, and the integral is strictly bigger than the integral #\int_{1}^{\infty}\frac{1}{x}\, \dd x#. If we calculate this integral, we get
\[\int_1^{\infty}\frac{1}{x}\, \dd x= \left[\ln \abs{x}\right]_{1}^{\infty}=\ln \abs{\infty}-\ln \abs{1} = \ln \abs{\infty} = \infty \]
Since #\ln \abs{x}# increases if #x# increases, we can say that #\ln \abs{\infty}# is equal to infinity. This is actually very contradictory, we have horn with a finite volume but with an infinite surface. If we would want to paint the surface of the horn, all the paint that fits in the volume of the horn would not be enough, even while this paint is also on the inside surface of the horn?
This quirky apparently contradictory problem was invented in the 17th century by the Italian scientist Evangelista Torricelli, and can therefore also be called the Trumpet of Torricelli.
Just as with areas we can also calculate the volume of a body resulting by revolving a surface between two graphs around the #x#-axis.
The surface #S# is enclosed by the functions #\blue f#, #\green{g}# with #\blue{f} \gt \green{g}# and the lines #x=a# and #x=b#.
The volume of the body #\orange B#, which is created if #V# revolves around the #x#-axis is equal to:
\[V(\orange B)=\pi \cdot \int_a^b (\blue{f(x)})^2-(\green{g(x)})^2 \; \dd x\]
We can prove this by subtracting the volume of the solid of revolution around the #x#-axis below #\green{g}# from the solid of revolution around the #x#-axis below#\blue{f}#.
\[\pi \int_a^b (\blue{f(x)})^2 \; \dd x-\pi \int_a^b (\green{g(x)})^2 \; \dd x =\pi \cdot \int_a^b (\blue{f(x)})^2-(\green{g(x)})^2 \; \dd x\]
Up till now we have seen how to revolve a surface around the #x#-axis, but we can also revolve a surface around the #y#-axis.
The surface #S# is enclosed by the function #\blue{f(x)}#, the #y#-axis and the lines #y=a# and #y=b#.
The volume of the body #\orange B#, which is created if #S# revolves around the #y#-axis is equal to
\[V(\orange B)=\pi \cdot \int_a^b x^2 \; \dd y\]
Note that we have to express #x# in the integral in #y# by means of #\blue{f(x)}#.
We prove this in the same way as with a solid of revolution around the #x#-axis. Hence, we will show that the derivative of the volume is equal to #\pi (x)^2#, since then #V(y)# is a primitive of #\pi (x)^2#.
For #V'(y)# according to the definition of the derivative, we have
\[V'(y) = \frac{V(y+h)-V(y)}{h} \text{ with } h \to 0\]
For small #h# we can approach #V(y+h)-V(y)# by a cilinder with ground surface with area #\pi x^2# and height #h#. Hence
\[V(y+h)-V(y)= \pi x^2 \cdot h\]
This means that
\[\frac{V(y+h)-V(y)}{h}=\pi x^2\]
Hence, like we wanted to prove, #V'(y)=\pi x^2#. Therefore we have #V(y)=\pi \int x^2 \; \dd y#.
The surface #S# is enclosed by the function #f(x)={{x^2}\over{2}}+4#, the #x#-axis and the lines #x=2# and #x=6#.
Body #B# is created by revolving surface #S# around the #x#-axis.
Calculate the volume of the body #B#. Simplify your answer as much as possible.
#V(B)=# #{{10928\cdot \pi}\over{15}}#
#\begin{array}{rcl}V(B)&=& \displaystyle \pi \cdot \int_{2}^{6} \left(f(x)\right)^2 \; \dd x \\ &&\phantom{xxx}\blue{\text{formula for volume solid of revolution}} \\ &=& \displaystyle \pi \cdot \int_{2}^{6} \left({{x^2}\over{2}}+4\right)^2 \; \dd x \\ &&\phantom{xxx}\blue{f(x)={{x^2}\over{2}}+4 \text{ entered}} \\&=& \displaystyle \pi \cdot \int_{2}^{6} {{x^4}\over{4}}+4\cdot x^2+16 \; \dd x \\ &&\phantom{xxx}\blue{\text{square expanded}} \\ &=& \displaystyle \pi \cdot \left[{{x^5}\over{20}}+{{4\cdot x^3}\over{3}}+16\cdot x\right]_{2}^{6} \\ &&\phantom{xxx}\blue{\text{taken the anti-derivative}} \\ &=& \displaystyle \pi\left({{3864}\over{5}}-{{664}\over{15}}\right) \\ &&\phantom{xxx}\blue{\text{boundaries entered}} \\ &=& \displaystyle {{10928\cdot \pi}\over{15}} \\ &&\phantom{xxx}\blue{\text{calculated}} \end{array}#