# Calculus/Volume of solids of revolution/Solutions

1. Calculate the volume of the cone with radius $r$ and height $h$ which is generated by the revolution of the region bounded by $y=r-{\frac {r}{h}}x$ and the lines $y=0$ and $x=0$ around the $x$ -axis.
The region extends in the $x$ -direction from $x=0$ to $x=h$ . The volume of the solid of revolution is given by
{\begin{aligned}\int _{0}^{h}\pi \left(r-{\frac {r}{h}}x\right)^{2}dx&=\pi \int _{0}^{h}\left(r^{2}-{\frac {2r^{2}}{h}}x+{\frac {r^{2}}{h^{2}}}x^{2}\right)dx\\&=\pi r^{2}\left(\int _{0}^{h}dx-{\frac {2}{h}}\int _{0}^{h}xdx+{\frac {1}{h^{2}}}\int _{0}^{h}x^{2}dx\right)\\&=\pi r^{2}\left(x{\bigr |}_{0}^{h}-{\frac {x^{2}}{h}}{\Bigr |}_{0}^{h}+{\frac {x^{3}}{3h^{2}}}{\Bigr |}_{0}^{h}\right)\\&=\pi r^{2}\left(h-h+{\frac {h}{3}}\right)\\&=\mathbf {\frac {\pi r^{2}h}{3}} \end{aligned}} The region extends in the $x$ -direction from $x=0$ to $x=h$ . The volume of the solid of revolution is given by
{\begin{aligned}\int _{0}^{h}\pi \left(r-{\frac {r}{h}}x\right)^{2}dx&=\pi \int _{0}^{h}\left(r^{2}-{\frac {2r^{2}}{h}}x+{\frac {r^{2}}{h^{2}}}x^{2}\right)dx\\&=\pi r^{2}\left(\int _{0}^{h}dx-{\frac {2}{h}}\int _{0}^{h}xdx+{\frac {1}{h^{2}}}\int _{0}^{h}x^{2}dx\right)\\&=\pi r^{2}\left(x{\bigr |}_{0}^{h}-{\frac {x^{2}}{h}}{\Bigr |}_{0}^{h}+{\frac {x^{3}}{3h^{2}}}{\Bigr |}_{0}^{h}\right)\\&=\pi r^{2}\left(h-h+{\frac {h}{3}}\right)\\&=\mathbf {\frac {\pi r^{2}h}{3}} \end{aligned}} 2. Calculate the volume of the solid of revolution generated by revolving the region bounded by the curve $y=x^{2}$ and the lines $x=1$ and $y=0$ around the $x$ -axis.
The region extends in the $x$ -direction from $x=0$ to $x=1$ . The volume of the solid of revolution is given by
{\begin{aligned}\int _{0}^{1}\pi \left(x^{2}\right)^{2}dx&=\pi \int _{0}^{1}x^{4}dx\\&=\pi {\frac {x^{5}}{5}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{5}} \end{aligned}} The region extends in the $x$ -direction from $x=0$ to $x=1$ . The volume of the solid of revolution is given by
{\begin{aligned}\int _{0}^{1}\pi \left(x^{2}\right)^{2}dx&=\pi \int _{0}^{1}x^{4}dx\\&=\pi {\frac {x^{5}}{5}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{5}} \end{aligned}} 3. Use the washer method to find the volume of a cone containing a central hole formed by revolving the region bounded by $y=R-{\frac {R}{h}}x$ and the lines $y=r$ and $x=0$ around the $x$ -axis.
The $x$ values of the region extend from $x=0$ to $x=h$ . The volume is
{\begin{aligned}V&=\int _{0}^{h}\pi \left(\left(R-{\frac {R}{h}}x\right)^{2}-r^{2}\right)dx\\&=\int _{0}^{h}\pi \left(R^{2}-2{\frac {R^{2}}{h}}x+{\frac {R^{2}}{h^{2}}}x^{2}-r^{2}\right)dx\\&=\pi \left(R^{2}x-{\frac {R^{2}}{h}}x^{2}+{\frac {R^{2}}{3h^{2}}}x^{3}-r^{2}x\right){\biggr |}_{0}^{h}\\&=\pi \left(R^{2}h-R^{2}h+{\frac {R^{2}h}{3}}-r^{2}h\right)\\&=\mathbf {\pi h\left({\frac {R^{2}}{3}}-r^{2}\right)} \end{aligned}} The $x$ values of the region extend from $x=0$ to $x=h$ . The volume is
{\begin{aligned}V&=\int _{0}^{h}\pi \left(\left(R-{\frac {R}{h}}x\right)^{2}-r^{2}\right)dx\\&=\int _{0}^{h}\pi \left(R^{2}-2{\frac {R^{2}}{h}}x+{\frac {R^{2}}{h^{2}}}x^{2}-r^{2}\right)dx\\&=\pi \left(R^{2}x-{\frac {R^{2}}{h}}x^{2}+{\frac {R^{2}}{3h^{2}}}x^{3}-r^{2}x\right){\biggr |}_{0}^{h}\\&=\pi \left(R^{2}h-R^{2}h+{\frac {R^{2}h}{3}}-r^{2}h\right)\\&=\mathbf {\pi h\left({\frac {R^{2}}{3}}-r^{2}\right)} \end{aligned}} 4. Calculate the volume of the solid of revolution generated by revolving the region bounded by the curves $y=x^{2}$ and $y=x^{3}$ and the lines $x=1$ and $y=0$ around the $x$ -axis.
:{\begin{aligned}V&=\int _{0}^{1}\pi \left(\left(x^{2}\right){}^{2}-\left(x^{3}\right){}^{2}\right)dx\\&=\int _{0}^{1}\pi \left(x^{4}-x^{6}\right)dx\\&=\pi \left({\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}\right){\biggr |}_{0}^{1}\\&=\pi \left({\frac {1}{5}}-{\frac {1}{7}}\right)\\&=\pi {\frac {7-5}{35}}\\&=\mathbf {\frac {2\pi }{35}} \end{aligned}} :{\begin{aligned}V&=\int _{0}^{1}\pi \left(\left(x^{2}\right){}^{2}-\left(x^{3}\right){}^{2}\right)dx\\&=\int _{0}^{1}\pi \left(x^{4}-x^{6}\right)dx\\&=\pi \left({\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}\right){\biggr |}_{0}^{1}\\&=\pi \left({\frac {1}{5}}-{\frac {1}{7}}\right)\\&=\pi {\frac {7-5}{35}}\\&=\mathbf {\frac {2\pi }{35}} \end{aligned}} 5. Find the volume of a cone with radius $R$ and height $h$ by using the shell method on the appropriate region which, when rotated around the $y$ -axis, produces a cone with the given characteristics.
You could set up the appropriate region in any of the four quadrants. Here we set it up in the first quadrant. Since we are revolving around the $y$ -axis, the $y$ direction will be the height and the radius will be along the $x$ direction. So we need a line that passes through the points $(0,h)$ and $(R,0)$ . The slope of this line is
$m={\frac {0-h}{R-0}}=-{\frac {h}{R}}$ and the $y$ -intercept is $h$ . Thus, the equation of the line is

$y=-{\frac {h}{R}}x+h$ The $x$ -values of the region run from $x=0$ to $x=R$ . Since the function is positive throughout the region we can drop the absolute value sign. The volume will be

{\begin{aligned}V&=\int _{0}^{R}2\pi x(-{\frac {h}{R}}x+h)dx\\&=\int _{0}^{R}2\pi (-{\frac {h}{R}}x^{2}+hx)dx\\&=2\pi (-{\frac {h}{3R}}x^{3}+{\frac {h}{2}}x^{2}){\biggr |}_{0}^{R}\\&=2\pi (-{\frac {hR^{2}}{3}}+{\frac {hR^{2}}{2}})\\&=2\pi {\frac {-2hR^{2}+3hR^{2}}{6}}\\&=\mathbf {\frac {\pi hR^{2}}{3}} \end{aligned}} You could set up the appropriate region in any of the four quadrants. Here we set it up in the first quadrant. Since we are revolving around the $y$ -axis, the $y$ direction will be the height and the radius will be along the $x$ direction. So we need a line that passes through the points $(0,h)$ and $(R,0)$ . The slope of this line is
$m={\frac {0-h}{R-0}}=-{\frac {h}{R}}$ and the $y$ -intercept is $h$ . Thus, the equation of the line is

$y=-{\frac {h}{R}}x+h$ The $x$ -values of the region run from $x=0$ to $x=R$ . Since the function is positive throughout the region we can drop the absolute value sign. The volume will be

{\begin{aligned}V&=\int _{0}^{R}2\pi x(-{\frac {h}{R}}x+h)dx\\&=\int _{0}^{R}2\pi (-{\frac {h}{R}}x^{2}+hx)dx\\&=2\pi (-{\frac {h}{3R}}x^{3}+{\frac {h}{2}}x^{2}){\biggr |}_{0}^{R}\\&=2\pi (-{\frac {hR^{2}}{3}}+{\frac {hR^{2}}{2}})\\&=2\pi {\frac {-2hR^{2}+3hR^{2}}{6}}\\&=\mathbf {\frac {\pi hR^{2}}{3}} \end{aligned}} 6. Calculate the volume of the solid of revolution generated by revolving the region bounded by the curve $y=x^{2}$ and the lines $x=1$ and $y=0$ around the $y$ -axis.
:{\begin{aligned}V&=\int _{0}^{1}2\pi xx^{2}dx\\&=\int _{0}^{1}2\pi x^{3}dx\\&=2\pi {\frac {x^{4}}{4}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{2}} \end{aligned}} :{\begin{aligned}V&=\int _{0}^{1}2\pi xx^{2}dx\\&=\int _{0}^{1}2\pi x^{3}dx\\&=2\pi {\frac {x^{4}}{4}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{2}} \end{aligned}} 