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# Calculus of Parametric Equations

Parametric equations can be quite handy, and we don't want to unravel them just to do Calculus. Some tricks can bend traditional derivative and integral methods to apply to parametric equations.

# Parametric Equations - Volume

Find the volume of the three dimensional shape obtained by revolving the curve described by the parametric equations below about the $$x$$-axis from $$t=0$$ to $$\displaystyle t=\frac{\pi}{2}:$$ \begin{align} x(t) &= 3 \cos t \\ y(t) &= 4 \sin t. \end{align}

Find the volume of the three dimensional shape obtained by revolving the curve described by the parametric equations below about the $$y$$-axis from $$t=1$$ to $$t=5:$$ \begin{align} x(t) &= 15 t^2 \\ y(t) &= 6 t. \end{align}

Find the volume of the three dimensional shape obtained by the following equations from $$r=0$$ to $$r= 1,$$ from $$t=0$$ to $$t= 3,$$ and $$u = 0$$ to $$u = 2 \pi:$$ \begin{align} x(r,t,u) &= 3 t \\ y(r,t,u) &= r\sqrt{ t^2+ 6t } \cos u \\ z(r,t,u) &= r\sqrt{ t^2 + 6t } \sin u. \end{align}

Find the volume of the three dimensional shape obtained by the following equations from $$t=-5$$ to $$t= 5,$$ $$u=-4$$ to $$u= 4,$$ and $$w = 0$$ to $$w = 1:$$ \begin{align} x(t,u,w) &= (1-w)t \\ y(t,u,w) &= (1-w)u \\ z(t,u,w) &= 9 w. \end{align}

Find the volume of the three dimensional shape obtained by the following equations from $$r = 0$$ to $$r = 1,$$ from $$t=0$$ to $$t=3 \pi ,$$ and $$u = 0$$ to $$u = 2 \pi:$$ \begin{align} x(r,t,u) &= 9 rt ( \cos u ) \\ y(r,t,u) &= 2 rt ( \sin u) \\ z(r,t,u) &= 4t. \end{align}

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