Volume of a Solid of Revolution: Disks and Washers
Solved Problems
Example 5.
A symmetrical parabolic segment with base \(a\) and height \(h\) rotates around the base. Calculate the volume of the resulting solid of revolution (Cavalieri's "lemon" ).
Solution.
The quadratic function is defined by the equation \(y = kx\left( {a - x} \right),\) where the coefficient \(k\) can be found from the condition \(y\left( {\frac{a}{2}} \right) = h.\) Hence
\[y\left( {\frac{a}{2}} \right) = h,\;\; \Rightarrow \frac{{ka}}{2}\left( {a - \frac{a}{2}} \right) = h,\;\; \Rightarrow \frac{{k{a^2}}}{4} = h,\;\; \Rightarrow k = \frac{{4h}}{{{a^2}}}.\]
So the parabolic segment is given by the expression \(y = \frac{{4h}}{{{a^2}}} x\left( {a - x} \right).\)
Figure 9. Using the disk method, we obtain
\[V = \pi \int\limits_0^a {{{\left[ {y\left( x \right)} \right]}^2}dx} = \pi \int\limits_0^a {{{\left[ {\frac{{4h}}{{{a^2}}}x\left( {a - x} \right)} \right]}^2}dx} = \frac{{16\pi {h^2}}}{{{a^4}}}\int\limits_0^a {{{\left( {ax - {x^2}} \right)}^2}dx} = \frac{{16\pi {h^2}}}{{{a^4}}}\int\limits_0^a {\left( {{a^2}{x^2} - 2a{x^3} + {x^4}} \right)dx} = \frac{{16\pi {h^2}}}{{{a^4}}}\left. {\left( {\frac{{{a^2}{x^3}}}{3} - \frac{{2a{x^4}}}{4} + \frac{{{x^5}}}{5}} \right)} \right|_0^a = \frac{{16\pi {h^2}}}{{{a^4}}}\left( {\frac{{{a^5}}}{3} - \frac{{{a^5}}}{2} + \frac{{{a^5}}}{5}} \right) = 16\pi {h^2}a \cdot \frac{1}{{30}} = \frac{{8\pi {h^2}a}}{{15}}.\]
Example 6.
The catenary line \[y = \cosh x\] rotates around the \(x-\)axis and produces a surface called a catenoid . Find the volume of the solid bounded by the catenoid and two planes \(x = -1\) and \(x = 1.\)
Solution.
Figure 10. Using the disk method and the hyperbolic identity
\[{\cosh ^2}x = \frac{{1 + \cosh 2x}}{2},\]
we have
\[V = \pi \int\limits_{ - 1}^1 {{{\left[ {\cosh x} \right]}^2}dx} = \frac{\pi }{2}\int\limits_{ - 1}^1 {\left[ {1 + \cosh 2x} \right]dx} = \frac{\pi }{2}\left. {\left[ {x + \frac{{\sinh 2x}}{2}} \right]} \right|_{ - 1}^1 = \frac{\pi }{2}\left[ {\left( {1 + \frac{{\sinh 2}}{2}} \right) - \left( { - 1 + \frac{{\sinh \left( { - 2} \right)}}{2}} \right)} \right] = \frac{\pi }{2}\left( {\sinh 2 + 2} \right).\]
We took into account here that the hyperbolic sine function is odd, so \(\sinh \left( { - 2} \right) = - \sinh 2.\)
Example 7.
Find the volume of the solid obtained by rotating the astroid \[x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1\] around its axis of symmetry.
Solution.
Due to symmetry, we can consider the region lying in the first quadrant and then multiply the volume of the region by two.
Figure 11. Let's solve the astroid equation for \({y^2}:\)
\[{x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = 1,\;\; \Rightarrow {y^{\frac{2}{3}}} = 1 - {x^{\frac{2}{3}}},\;\; \Rightarrow {\left( {{y^{\frac{2}{3}}}} \right)^3} = {\left( {1 - {x^{\frac{2}{3}}}} \right)^3},\;\; \Rightarrow {y^2} = 1 - 3{x^{\frac{2}{3}}} + 3{x^{\frac{4}{3}}} - {x^2}.\]
Hence, the total volume of the solid bounded by the astroid is given by
\[V = 2\pi \int\limits_0^1 {{y^2}\left( x \right)dx} = 2\pi \int\limits_0^1 {\left( {1 - 3{x^{\frac{2}{3}}} + 3{x^{\frac{4}{3}}} - {x^2}} \right)dx} = 2\pi \left. {\left( {x - \frac{{9{x^{\frac{5}{3}}}}}{5} + \frac{{9{x^{\frac{7}{3}}}}}{7} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = 2\pi \left( {1 - \frac{9}{5} + \frac{9}{7} - \frac{1}{3}} \right) = 2\pi \cdot \frac{{16}}{{105}} = \frac{{32\pi }}{{105}}.\]
Example 8.
Calculate the volume of the solid obtained by rotating the region bounded by the curve \[y = 2x - {x^2}\] and the \(x-\)axis about the \(y-\)axis.
Solution.
Find the points of intersection of the parabola with the \(x-\)axis:
\[2x - {x^2} = 0,\;\; \Rightarrow x\left( {2 - x} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = 2.\]
As the region is revolved about the \(y-\)axis, we express the equation of the bounding curve in terms of \(y:\)
\[y = 2x - {x^2},\;\; \Rightarrow {x^2} - 2x + 1 = 1 - y,\;\; \Rightarrow {\left( {x - 1} \right)^2} = 1 - y,\;\; \Rightarrow x - 1 = \pm \sqrt {1 - y} ,\;\; \Rightarrow x = 1 \pm \sqrt {1 - y} .\]
The signs "plus" and "minus" correspond to the two branches of the curve:
\[x = g\left( y \right) = 1 - \sqrt {1 - y} ,\]
\[x = f\left( y \right) = 1 + \sqrt {1 - y} .\]
Figure 12. Given that the variable \(y\) varies from \(0\) to \(1\) and using the washer method, we find the volume of the solid:
\[V = \pi \int\limits_0^1 {\left( {{{\left[ {f\left( y \right)} \right]}^2} - {{\left[ {g\left( y \right)} \right]}^2}} \right)dy} = \pi \int\limits_0^1 {\left( {{{\left[ {1 + \sqrt {1 - y} } \right]}^2} - {{\left[ {1 - \sqrt {1 - y} } \right]}^2}} \right)dy} = \pi \int\limits_0^1 {\left( {4\sqrt {1 - y} } \right)dy} = 4\pi \int\limits_0^1 {\sqrt {1 - y} dy} = \left. {\left[ {4\pi \cdot \frac{{2{{\left( {1 - y} \right)}^{\frac{3}{2}}}}}{3} \cdot \left( { - 1} \right)} \right]} \right|_0^1 = \left. {\left[ { - \frac{{8\pi \sqrt {{{\left( {1 - y} \right)}^3}} }}{3}} \right]} \right|_0^1 = \frac{{8\pi }}{3}.\]
Example 9.
Find the volume of the solid obtained by rotating an equilateral triangle with side \(a\) around one of its sides.
Solution.
Let \(ABC\) be an equilateral triangle with side \(a\).
Figure 13. Determine the height of the triangle \(OB:\)
\[OB = \sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}} = \sqrt {\frac{{3{a^2}}}{4}} = \frac{{\sqrt 3 a}}{2}.\]
So, the vertices \(A,\) \(B,\) \(C\) have the following coordinates:
\[A\left( {0,\frac{a}{2}} \right),\;B\left( {\frac{{\sqrt 3 a}}{2},0} \right),\;C\left( {0, - \frac{a}{2}} \right).\]
Find the equation of the straight line \(AB\) using the two-point form:
\[\frac{{x - {x_A}}}{{{x_B} - {x_A}}} = \frac{{y - {y_A}}}{{{y_B} - {y_A}}},\;\; \Rightarrow \frac{{x - 0}}{{\frac{{\sqrt 3 a}}{2} - 0}} = \frac{{y - \frac{a}{2}}}{{0 - \frac{a}{2}}},\;\; \Rightarrow \frac{x}{{\sqrt 3 }} = \frac{{y - \frac{a}{2}}}{{ - 1}}, \;\;\Rightarrow x = \frac{{a\sqrt 3 }}{2} - \sqrt 3 y.\]
By symmetry of the solid, we can integrate from \(0\) to \(\frac{a}{2}\) and then multiply the result by two. Hence, the volume is given by the formula:
\[V = 2\pi \int\limits_0^{\frac{a}{2}} {{{\left[ {x\left( y \right)} \right]}^2}dy} = 2\pi \int\limits_0^{\frac{a}{2}} {{{\left[ {\frac{{a\sqrt 3 }}{2} - \sqrt 3 y} \right]}^2}dy} = 6\pi \int\limits_0^{\frac{a}{2}} {{{\left[ {\frac{a}{2} - y} \right]}^2}dy} = 6\pi \int\limits_0^{\frac{a}{2}} {\left[ {\frac{{{a^2}}}{4} - ay + {y^2}} \right]dy} = 6\pi \left. {\left[ {\frac{{{a^2}y}}{4} - \frac{{a{y^2}}}{2} + \frac{{{y^3}}}{3}} \right]} \right|_0^{\frac{a}{2}} = 6\pi \left[ {\cancel{\frac{{{a^3}}}{8}} - \cancel{\frac{{{a^3}}}{8}} + \frac{{{a^3}}}{{24}}} \right] = \frac{{6\pi {a^3}}}{{24}} = \frac{{\pi {a^3}}}{4}.\]
Example 10.
One arch of the cycloid \[x = \theta - \sin \theta, y = 1 - \cos \theta\] revolves around its base. Calculate the volume of the body bounded by the given surface.
Solution.
Figure 14. The cycloid is given in parametric form. Therefore we express the integral \(V = \pi\int\limits_0^{2\pi } {{y^2}dx} \) in terms of the parameter \(\theta:\)
\[{y^2} = {\left( {1 - \cos \theta } \right)^2} ,\]
\[dx = d\left( {\theta - \sin \theta } \right) = \left( {1 - \cos \theta } \right)d\theta .\]
Note that the variable \(x\) and the parameter \(\theta\) change in the same range from \(0\) to \(2\pi.\) Hence, the volume of the solid is given by the integral
\[V = \pi \int\limits_0^{2\pi } {{y^2}dx} = \pi \int\limits_0^{2\pi } {{{\left( {1 - \cos \theta } \right)}^3}d\theta } .\]
To calculate the integral we use the following algebraic and trigonometric identities:
\[\left( {a - b} \right)^3 = {a^3} - 3{a^2}b + 3a{b^2} - {b^3},\]
\[{\cos ^2}\theta = \frac{1}{2} + \frac{1}{2}\cos 2\theta ,\]
\[{\cos ^3}\theta = \frac{3}{4}\cos \theta + \frac{1}{4}\cos 3\theta .\]
Hence, the volume of the solid is
\[V = \pi \int\limits_0^{2\pi } {{{\left( {1 - \cos \theta } \right)}^3}d\theta } = \pi \int\limits_0^{2\pi } {\left( {1 - 3\cos \theta + 3\,{{\cos }^2}\theta - {{\cos }^3}\theta } \right)d\theta } = \pi \int\limits_0^{2\pi } {\left( {1 - 3\cos \theta + \frac{3}{2} + \frac{3}{2}\cos 2\theta - \frac{3}{4}\cos \theta - \frac{1}{4}\cos 3\theta } \right)d\theta } = \pi \int\limits_0^{2\pi } {\left( {\frac{5}{2} - \frac{{15}}{4}\cos \theta + \frac{3}{2}\cos 2\theta - \frac{1}{4}\cos 3\theta } \right)d\theta } = \pi \left. {\left[ {\frac{{5\theta }}{2} - \frac{{15}}{4}\sin \theta + \frac{3}{4}\sin 2\theta - \frac{1}{{12}}\sin 3\theta } \right]} \right|_0^{2\pi } = 5{\pi ^2}.\]
