# Thread: cycloid parametrically and surface area

1. ## cycloid parametrically and surface area

Hi I'm new here, this is my first post I hope you don't mind if I ask 2 questions I'm stuck at.

1. a cycloid is given parametrically by x = t-sint, y= 1 - cost
find the arc length between t = 0 and t = 2 pie

2. find the surface area when the curve y = (square root of)2x + 1 x E [1, 7] is rotated by 360 degrees about the x axis

2. Hello, graeme87!

Welcome aboard!
Here's the first one . . .

1. A cycloid is given parametrically by: . $\begin{array}{ccc} x &=& t-\sin t \\ y&=& 1 - \cos t\end{array}$

Find the arc length between $t = 0$ and $t = 2\pi$
Formula: . $L \;=\;\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \,dt$

We have: . $\begin{array}{ccc}\dfrac{dx}{dt} &=& 1 -\cos t \\ \\[-3mm] \dfrac{dy}{dt} &=& \sin t \end{array}$

$\text{Then: }\;\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \;=\;(1\;-\;\cos t)^2 + \;\sin^2\!t \quad\Rightarrow\quad 1 \;-\; 2\cos t \;+\; \underbrace{\cos^2\!t + \sin^2\!t }_{\text{This is 1}}$

. . . . . . $= \;2 - 2\cos t \;=\;2(1 - \cos t) \;=\;4\left(\frac{1-\cos t}{2}\right) \;=\;4\sin^2\tfrac{t}{2}$

Hence: . $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \;=\;2\sin\tfrac{t}{2}$

We have: . $L \;=\;\int^{2\pi}_0 \!\!2\sin\tfrac{t}{2}\,dt \;=\;-4\cos\tfrac{t}{2}\,\bigg]^{2\pi}_0 \;=\;\bigg(-4\cos\pi\bigg) - \bigg(-4\cos 0\bigg)$

. . . . . . $= \;(\text{-}4)(\text{-}1) - (\text{-}4) \;=\; 4 + 4 \;=\;\boxed{8}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

That is a surprising answer . . .

We have a circle . . . and we are rolling it.

We would expect to see $\pi$ in the arc length.

Fact: The arc length of one arch of a cycloid is equal to the perimeter
. . . . of the square circumscribed about the generating circle.

Fact: The area under one arch of a cycloid
. . . . is 3 times the area of the generating circle.
Code:
                        * * *...
*  *           *::*..
*     *               *:::::*
*        *                 *::::::::*
*                              :::::::::::*
*            *        πrē        *::::::::::::*
πrē                         ::::: πrē::::::
*               *                 *:::::::::::::::*
*               *::::::::::::::::::
*           *::::::::::::::::::::
* - - - - - - - - - - - * * * - - - - - - - - - - - *

That shark-fin-shaped region is equal to the generating circle.

3. Thank you, you're help is very much appreciated!