MHB 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 $$r = cos(2\theta)$$.

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

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

 

[shamieh]

So would the first step be to solve for $$\theta$$?

So:

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

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

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

then $$\theta = \frac{cos^{-1}(r)}{2}$$ ? 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 $$r = cos(2\theta)$$.

I know that the formula is $$A = \int ^{\beta}_{\alpha} \frac{1}{2} [f(\theta)]^2\color{red}{d\theta}$$ 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:

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

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

Finding everywhere that $$cos = 0, $$

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

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

$$\therefore$$ I can say: $$2\theta = \frac{\pi}{2} = \frac{\pi}{4} $$

similarly $$2\theta = \frac{3\pi}{2} = \frac{3\pi}{4}$$

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

 

[Opalg]

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

So:

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

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

Finding everywhere that $$cos = 0, $$

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

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

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

similarly $$\color{red}{2\theta = \frac{3\pi}{2} = \frac{3\pi}{4}}$$

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

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 $$2\theta = \frac{\pi}{2}$$

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

we can then find $$\theta = \frac{\pi}{4}$$

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