# Thread: volume of parametric equation

1. ## volume of parametric equation

Find the volume of the solid formed when one arch of the cycloid defined parametrically by $x=\theta-sin(\theta)$ and $y=1-cos(\theta)$ is rotated about the x-axis.

2. What I would do is plug in a few points for both these functions using the unit circle for theta. From there, you can plug in those (x,y) coordinates to your graphing calculator and run exponential and sin regression tests. Then, just use the function with the |r| value closest to 1 and find volume as you normally would.

3. Originally Posted by macosxnerd101
What I would do is plug in a few points for both these functions using the unit circle for theta. From there, you can plug in those (x,y) coordinates to your graphing calculator and run exponential and sin regression tests. Then, just use the function with the |r| value closest to 1 and find volume as you normally would.
how do you do the exponential and sin regression tests?

4. To be perfectly honest, I think macosxnerd101 is just pulling your leg!

Here's what i would do: Write the surface in terms of the $\theta$ and $\phi$ parameters where $\phi$ is the angle of rotation. Rotating around the x-axis, the radius of each circle is just the y coordinate. $x= \theta- sin(\theta)$ , $y= (1- cos(\theta))cos(\phi)$, and $z= (1- cos(\theta))sin(\phi)$.

Now we can write the "position vector" for each point on the surface:
$\vec{r}(\theta, \phi)= x\vec{i}+ yvec{j}+ z\vec{k}= (\theta- sin(\theta))\vec{i}+ (1- cos(\theta))cos(\phi)\vec{j}+ (1- cos(\theta))sin(\phi)\vec{k}$.

The dervatives with respect to the parameters:
$\vec{r}_\theta= (1- cos(\theta))\vec{i}+ sin(\theta)cos(\phi)\vec{j}+ sin(\theta)sin(\phi)\vec{k}$ and
$\vec{r}_\phi= -(1- cos(\theta))sin(\phi)\vec{j}+ (1- cos(\theta))cos(\phi)\vec{k}$

Are tangent vectors to the surface and their cross product
$sin(\theta)(1- cos(\theta))\vec{i}- (1- cos(\theta))^2cos(\phi)\vec{j}- (1- cos(\theta))^2sin(\phi)\vec{k}$

also called the "fundamental vector product" for the surface, is normal to the surface and its length
$sin(\theta)\sqrt{1+ (1- cos(\theta))^2}$

gives the "differential of surface area". That is, the surface area is
$\int_{\phi= 0}^{2\pi}\int_{\theta= 0}^{\pi} sin(\theta)\sqrt{1+ (1- cos(\theta))^2}d\theta d\phi$

The $\phi$ integral is trivial and I recommend the substitution $u= 1- cos(\theta)$ for the $\theta$ integral.