# Calculus/Volume of solids of revolution/Solutions

$r$
$h$
$y=r-{\frac {r}{h}}x$
$y=0$
$x=0$
$x$
$x$
$x=0$
$x=h$
    ${\begin{aligned}\int _{0}^{h}\pi \left(r-{\frac {r}{h}}x\right)^{2}dx&=\pi \int _{0}^{h}\left(r^{2}-{\frac {2r^{2}}{h}}x+{\frac {r^{2}}{h^{2}}}x^{2}\right)dx\\&=\pi r^{2}\left(\int _{0}^{h}dx-{\frac {2}{h}}\int _{0}^{h}xdx+{\frac {1}{h^{2}}}\int _{0}^{h}x^{2}dx\right)\\&=\pi r^{2}\left(x{\bigr |}_{0}^{h}-{\frac {x^{2}}{h}}{\Bigr |}_{0}^{h}+{\frac {x^{3}}{3h^{2}}}{\Bigr |}_{0}^{h}\right)\\&=\pi r^{2}\left(h-h+{\frac {h}{3}}\right)\\&=\mathbf {\frac {\pi r^{2}h}{3}} \end{aligned}}$
$x$
$x=0$
$x=h$
    ${\begin{aligned}\int _{0}^{h}\pi \left(r-{\frac {r}{h}}x\right)^{2}dx&=\pi \int _{0}^{h}\left(r^{2}-{\frac {2r^{2}}{h}}x+{\frac {r^{2}}{h^{2}}}x^{2}\right)dx\\&=\pi r^{2}\left(\int _{0}^{h}dx-{\frac {2}{h}}\int _{0}^{h}xdx+{\frac {1}{h^{2}}}\int _{0}^{h}x^{2}dx\right)\\&=\pi r^{2}\left(x{\bigr |}_{0}^{h}-{\frac {x^{2}}{h}}{\Bigr |}_{0}^{h}+{\frac {x^{3}}{3h^{2}}}{\Bigr |}_{0}^{h}\right)\\&=\pi r^{2}\left(h-h+{\frac {h}{3}}\right)\\&=\mathbf {\frac {\pi r^{2}h}{3}} \end{aligned}}$
$y=x^{2}$
$x=1$
$y=0$
$x$
$x$
$x=0$
$x=1$
    ${\begin{aligned}\int _{0}^{1}\pi \left(x^{2}\right)^{2}dx&=\pi \int _{0}^{1}x^{4}dx\\&=\pi {\frac {x^{5}}{5}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{5}} \end{aligned}}$
$x$
$x=0$
$x=1$
    ${\begin{aligned}\int _{0}^{1}\pi \left(x^{2}\right)^{2}dx&=\pi \int _{0}^{1}x^{4}dx\\&=\pi {\frac {x^{5}}{5}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{5}} \end{aligned}}$
$y=R-{\frac {R}{h}}x$
$y=r$
$x=0$
$x$
$x$
$x=0$
$x=h$
    ${\begin{aligned}V&=\int _{0}^{h}\pi \left(\left(R-{\frac {R}{h}}x\right)^{2}-r^{2}\right)dx\\&=\int _{0}^{h}\pi \left(R^{2}-2{\frac {R^{2}}{h}}x+{\frac {R^{2}}{h^{2}}}x^{2}-r^{2}\right)dx\\&=\pi \left(R^{2}x-{\frac {R^{2}}{h}}x^{2}+{\frac {R^{2}}{3h^{2}}}x^{3}-r^{2}x\right){\biggr |}_{0}^{h}\\&=\pi \left(R^{2}h-R^{2}h+{\frac {R^{2}h}{3}}-r^{2}h\right)\\&=\mathbf {\pi h\left({\frac {R^{2}}{3}}-r^{2}\right)} \end{aligned}}$
$x$
$x=0$
$x=h$
    ${\begin{aligned}V&=\int _{0}^{h}\pi \left(\left(R-{\frac {R}{h}}x\right)^{2}-r^{2}\right)dx\\&=\int _{0}^{h}\pi \left(R^{2}-2{\frac {R^{2}}{h}}x+{\frac {R^{2}}{h^{2}}}x^{2}-r^{2}\right)dx\\&=\pi \left(R^{2}x-{\frac {R^{2}}{h}}x^{2}+{\frac {R^{2}}{3h^{2}}}x^{3}-r^{2}x\right){\biggr |}_{0}^{h}\\&=\pi \left(R^{2}h-R^{2}h+{\frac {R^{2}h}{3}}-r^{2}h\right)\\&=\mathbf {\pi h\left({\frac {R^{2}}{3}}-r^{2}\right)} \end{aligned}}$
$y=x^{2}$
$y=x^{3}$
$x=1$
$y=0$
$x$
${\begin{aligned}V&=\int _{0}^{1}\pi \left(\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}\right)dx\\&=\int _{0}^{1}\pi \left(x^{4}-x^{6}\right)dx\\&=\pi \left({\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}\right){\biggr |}_{0}^{1}\\&=\pi \left({\frac {1}{5}}-{\frac {1}{7}}\right)\\&=\pi {\frac {7-5}{35}}\\&=\mathbf {\frac {2\pi }{35}} \end{aligned}}$
${\begin{aligned}V&=\int _{0}^{1}\pi \left(\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}\right)dx\\&=\int _{0}^{1}\pi \left(x^{4}-x^{6}\right)dx\\&=\pi \left({\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}\right){\biggr |}_{0}^{1}\\&=\pi \left({\frac {1}{5}}-{\frac {1}{7}}\right)\\&=\pi {\frac {7-5}{35}}\\&=\mathbf {\frac {2\pi }{35}} \end{aligned}}$
$R$
$h$
$y$
$y$
$y$
$x$
$(0,h)$
$(R,0)$
    $m={\frac {0-h}{R-0}}=-{\frac {h}{R}}$
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
    ${\begin{aligned}V&=\int _{0}^{R}2\pi x(-{\frac {h}{R}}x+h)dx\\&=\int _{0}^{R}2\pi (-{\frac {h}{R}}x^{2}+hx)dx\\&=2\pi (-{\frac {h}{3R}}x^{3}+{\frac {h}{2}}x^{2}){\biggr |}_{0}^{R}\\&=2\pi (-{\frac {hR^{2}}{3}}+{\frac {hR^{2}}{2}})\\&=2\pi {\frac {-2hR^{2}+3hR^{2}}{6}}\\&=\mathbf {\frac {\pi hR^{2}}{3}} \end{aligned}}$
$y$
$y$
$x$
$(0,h)$
$(R,0)$
    $m={\frac {0-h}{R-0}}=-{\frac {h}{R}}$
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
    ${\begin{aligned}V&=\int _{0}^{R}2\pi x(-{\frac {h}{R}}x+h)dx\\&=\int _{0}^{R}2\pi (-{\frac {h}{R}}x^{2}+hx)dx\\&=2\pi (-{\frac {h}{3R}}x^{3}+{\frac {h}{2}}x^{2}){\biggr |}_{0}^{R}\\&=2\pi (-{\frac {hR^{2}}{3}}+{\frac {hR^{2}}{2}})\\&=2\pi {\frac {-2hR^{2}+3hR^{2}}{6}}\\&=\mathbf {\frac {\pi hR^{2}}{3}} \end{aligned}}$
$y=x^{2}$
$x=1$
$y=0$
$y$
${\begin{aligned}V&=\int _{0}^{1}2\pi xx^{2}dx\\&=\int _{0}^{1}2\pi x^{3}dx\\&=2\pi {\frac {x^{4}}{4}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{2}} \end{aligned}}$
${\begin{aligned}V&=\int _{0}^{1}2\pi xx^{2}dx\\&=\int _{0}^{1}2\pi x^{3}dx\\&=2\pi {\frac {x^{4}}{4}}{\biggr |}_{0}^{1}\\&=\mathbf {\frac {\pi }{2}} \end{aligned}}$
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