Finding the area inside of a loop

[shamieh]

I know this is relatively easy but I'm just confused on the process...

Find the area inside one loop of a four leafed rose \(\displaystyle r = cos(2\theta)\).

I know that the formula is \(\displaystyle A = \int ^{\beta}_{\alpha} \frac{1}{2} [f(\theta)]^2\) right?

I'm just not sure what to plug in or solve for.

 

[shamieh]

So would the first step be to solve for \(\displaystyle \theta\)?

So:

\(\displaystyle r = cos(2\theta)\)

\(\displaystyle cos^{-1}(r) = cos^{-1}(cos(2\theta))\)

\(\displaystyle = cos^{-1}(r) = 2\theta\)

then \(\displaystyle \theta = \frac{cos^{-1}(r)}{2}\) ? But how does that help me?

 

[Opalg]

I know this is relatively easy but I'm just confused on the process...

Find the area inside one loop of a four leafed rose \(\displaystyle r = cos(2\theta)\).

I know that the formula is \(\displaystyle A = \int ^{\beta}_{\alpha} \frac{1}{2} [f(\theta)]^2\color{red}{d\theta}\) right? Dont't forget the $\color{red}{d\theta}$!

I'm just not sure what to plug in or solve for.

It will help to have an idea of what the graph looks like.

[graph]lfg70xit5x[/graph] (click on it for an enlargement).

The reason it has loops is that there are some values of $\theta$ for which $r=0.$ Start by finding those values. That will tell you what to take for $\alpha$ and $\beta$.

 

[shamieh]

I think I understand it now, can you verify if I have done this correctly?

So:

\(\displaystyle r = cos(2\theta)\)

I don't know what \(\displaystyle cos(2\theta)\) is, but I do know what \(\displaystyle cos(\theta) = 0\)

Finding everywhere that \(\displaystyle cos = 0, \)

I found that \(\displaystyle cos = 0 @ \frac{\pi}{2}\) and \(\displaystyle cos\) is \(\displaystyle 0 @ \frac{3\pi}{2}\)

But I need \(\displaystyle cos(2\theta)\)

\(\displaystyle \therefore\) I can say: \(\displaystyle 2\theta = \frac{\pi}{2} = \frac{\pi}{4} \)

similarly \(\displaystyle 2\theta = \frac{3\pi}{2} = \frac{3\pi}{4}\)

Thus my integral is \(\displaystyle \frac{1}{2} \int^{\frac{3\pi}{4}}_{\frac{\pi}{4}} cos^2(2\theta) \, d\theta\) and after skipping a few steps of taking the A.D. I obtained : \(\displaystyle \frac{\pi}{8}\)

 

[Opalg]

I think I understand it now, can you verify if I have done this correctly?

So:

\(\displaystyle r = cos(2\theta)\)

I don't know what \(\displaystyle cos(2\theta)\) is, but I do know what \(\displaystyle cos(\theta) = 0\)

Finding everywhere that \(\displaystyle cos = 0, \)

I found that \(\displaystyle cos = 0 @ \frac{\pi}{2}\) and \(\displaystyle cos\) is \(\displaystyle 0 @ \frac{3\pi}{2}\)

But I need \(\displaystyle cos(2\theta)\)

\(\displaystyle \therefore\) I can say: \(\displaystyle \color{red}{2\theta = \frac{\pi}{2} = \frac{\pi}{4}} \)

similarly \(\displaystyle \color{red}{2\theta = \frac{3\pi}{2} = \frac{3\pi}{4}}\)

Thus my integral is \(\displaystyle \frac{1}{2} \int^{\frac{3\pi}{4}}_{\frac{\pi}{4}} cos^2(2\theta) \, d\theta\) and after skipping a few steps of taking the A.D. I obtained : \(\displaystyle \frac{\pi}{8}\)

Your answer $\frac\pi8$ is correct. But your writing still shows some very bad habits. You can NOT say things like $\frac{\pi}{2} = \frac{\pi}{4}$ because it is blatantly untrue. You should only use the equals sign "=" to connect things that are equal to each other. What you should have written there is something like "$2\theta = \frac{\pi}{2}$ and therefore $\theta = \frac{\pi}{4}$".

 

[shamieh]

Thanks for the help. Found another way to solve this.

Setting \(\displaystyle 2\theta = \frac{\pi}{2}\)

Since we only need the first point at which \(\displaystyle \cos(\theta) = 0 \)

we can then find \(\displaystyle \theta = \frac{\pi}{4}\)

\(\displaystyle \therefore\) 2 [ \(\displaystyle \int ^{\frac{\pi}{4}}_0 \frac{1}{2} [r]^2 d\theta \) ] \(\displaystyle = \frac{\pi}{8}\)