area of a limacon

[jaggtagg7]

hm, can someone remind me of how you find the integration limits to find the area of the inner loop of a limacon?

the example equation we have is r=2-3sin(theta) and the limits are given as the lower being .7297 and the upper as 2.412. now i understand you set the equation equal to 0 to find the lower limit, but how do solve for the upper limit, as the equation only turns out the one answer for the lower.
thanks bunches.
christine

 

[soroban]

Hello, Christine!

We have: \(\displaystyle \,r\:=\:2\,-\,3\sin\theta\)
and the limits are given as: 0.7297 and 2.412.
Now i understand you set the equation equal to 0 to find the lower limit,
but how do solve for the upper limit as the equation only turns out the one answer for the lower.
Actually, that's not true . . .

We set the function equal to zero to find when the curve cross the pole (origin).

So we have: \(\displaystyle \:2\,-\,3\sin\theta\:=\:0\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{2}{3}\)

Maybe your <u>calculator</u> gives only one answer, but there are an inifnite number of solutions.

Recall that: \(\displaystyle \,\sin\theta\,=\,\frac{2}{3}\;\;\Rightarrow\;\;\theta\,=\, \arcsin\left(\frac{2}{3}\right)\:=\:\begin{Bmatrix}0.7297\,+\,2\pi n \\ 2.412\,+\,2\pi n\end{Bmatrix}\)


Recall that, if \(\displaystyle \sin\theta\,=\,\frac{1}{2}\), then: \(\displaystyle \theta\:=\:30^o\) or \(\displaystyle 150^o\) (and many others).

 

[jaggtagg7]

im still confused on how you solve for the value... or is it just a matter of knowing when the graph repeats? and how do you know when it repeats? the upper limit doesnt come from .7297+2(pi)n .. so how do you arrive at it?
thanks

 

[Gene]

What am I missing?
dA = r*d\(\displaystyle \Theta\)
A = \(\displaystyle 2\Theta-3cos\Theta\rbrac\)_.7297^2.412