Question

Find the area under one arch of the cycloid.

(still typing equation)

Answer 1

\[x=9(3\theta-\sin3\theta) \]
\[y=9(1-\cos3\theta)\]

Answer 2

(done typing equations)

Answer 3

it seems one arch corresponds to an angle of 2pi/3 ?

Answer 4

look at the lastg example
http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx

Answer 5

@iambatman

Answer 6

Ok so one arch of the cycloid is \[0 \le \theta \le 2\pi \]

now using the substitution rule with \[y=9(1-\cos 3 \theta) ~~~ and ~~~ dx = 9(1-\cos 3 \theta) d \theta\]

We then have \[A = \int\limits_{0}^{2 \pi r} y dx = \int\limits_{0}^{2\pi} 9(1-\cos 3 \theta)9(1-\cos 3 \theta) d \theta\]

Answer 7

The substitution rule is \[A = \int\limits_{a}^{b} y dx = \int\limits_{\alpha}^{\beta} g(t)f'(t) dt\]

right?

Answer 8

here period is 2pi/3 right

Answer 9

Yeah that's right

Answer 10

It's good to remember the period of the graph is \[\frac{ 2\pi }{ b }\] where b is a multiplicand of the angle inside the function.

Answer 11

@ganeshie8 Hey just to make sure so our interval would change right it would be then \[0 \le \theta \le \frac{ 2 \pi }{ 3 }\]

Answer 12

looks good to me!!