# Thread: Volume generated by curve in one complete rotation

1. ## Volume generated by curve in one complete rotation

Find the area bounded by the x-axis and the arc of the cycloid $\displaystyle x=a(\theta-\sin\theta)$, $\displaystyle y=a(1-\cos\theta)$ between $\displaystyle \theta =0$ and $\displaystyle \theta=2\pi$, rotated about the x-axis.

My problem is that I don't have the formula for the volume for parametric equations.
For Cartesian equations :$\displaystyle V=\int ^a_b (\pi y^2) dx$
But I don't know how to adapt this for parametric equations.
Thanks!

2. Hello, arze!

$\displaystyle \text{The region bounded by the }x\text{-axis and the arc of the cycloid:}$

. . $\displaystyle \begin{Bmatrix}x&=&a(\theta-\sin\theta) \\ y&=&a(1-\cos\theta)\end{Bmatrix}\:\text{ between }\theta =0\text{ and }\theta=2\pi$

$\displaystyle \text{is rotated about the }x\text{-axis.}$

My problem is that I don't have the formula for the volume for parametric equations.

For Cartesian equations: .$\displaystyle \displaystyle V\:=\:\pi \int ^a_b y^2\,dx$

But I don't know how to adapt this for parametric equations. . Yes, you do!

We have: .$\displaystyle \begin{Bmatrix}dx &=& a(1-\cos\theta)\,d\theta \\ y &=& a(1-\cos\theta) \end{Bmatrix}$

Substitute: .$\displaystyle \displaystyle V \;=\;\pi \int^b_a y^2\,dx$

. . . . . . . . . . . $\displaystyle \displaystyle =\;\pi \int^{2\pi}_0 \overbrace{a^2(1-\cos\theta)^2}^{y^2} \cdot \overbrace{a(1-\cos\theta)\,d\theta}^{dx}$

. . . . . . . . . . . $\displaystyle \displaystyle =\;\pi a^3 \int^{2\pi}_0 (1-\cos\theta)^3\,d\theta$

Got it?