Help With: Area (Polar Coordinates), Confusing Integral

[tolove]

Find the area of the following region:

Inside: r² = 6 cos 2θ
Outside: r = √3

Here's how I've set up the integral. I have to be making a mistake somewhere in the set up, but I can't figure it out.

r₁ = √3
r₂ = [itex](\sqrt{6 cos 2θ})[/itex]

[itex]\frac{Area}{4}[/itex]= [itex]\frac{1}{2}\int\ (\sqrt{6 cos 2θ})^{2} - \sqrt{3}^{2} dθ [/itex], Evaluated on θ = 0 .. [itex]\frac{π}{4}[/itex]

This is wrong, however. 3√3 - π is the correct solution to this problem. (If this is the correct setup, let me know and I'll post the steps to how I'm incorrectly solving it for help there.)

Also, the second thing I need help with, a frustrating integral:

6[itex]\int \frac{1}{(1+cos θ)^{2} } \sqrt{2 + 2cos θ} dθ[/itex]

Any hints on this? Thank you very much for reading!

 

[jbunniii]

Check your integration limits. The curve ##r^2 = 6 \cos 2\theta## is not the "inside" curve throughout the full range ##0 \leq \theta \leq \pi/4##. Try sketching both curves.

 

[tolove]

Check your integration limits. The curve ##r^2 = 6 \cos 2\theta## is not the "inside" curve throughout the full range ##0 \leq \theta \leq \pi/4##. Try sketching both curves.

Trust me I have!

http://www.wolframalpha.com/input/?i=+r2+=+6+cos+2θ

0 to π/4 is all the shading in the first quadrant, right? Can't I simply take that and multiply it by 4? I've also tried this from 0 .. π and I get the same wrong answer.

 

[jbunniii]

You need to identify the region where the curve ##r^2 = 6\cos 2\theta## is on the inside. That is true when
$$6\cos 2\theta \leq 3$$
or equivalently
$$\cos 2\theta \leq \frac{1}{2}$$
For what values of ##\theta## does this inequality hold?

 

[tolove]

You need to identify the region where the curve ##r^2 = 6\cos 2\theta## is on the inside. That is true when
$$6\cos 2\theta \leq 3$$
or equivalently
$$\cos 2\theta \leq \frac{1}{2}$$
For what values of ##\theta## does this inequality hold?

Ahh, I see. I was incorrectly assuming that pi/4 was the end point.

I needed to set the two r values together for their intercept point. 0 .. pi/6

Thank you very much!

 

[jbunniii]

Ahh, I see. I was incorrectly assuming that pi/4 was the end point.

I needed to set the two r values together for their intercept point. 0 .. pi/6

Thank you very much!

Well, it is true that ##\theta = \pi/6## is a point where the two curves meet. But for ##0 \leq \theta \leq \pi/6## the reverse inequality is true:
$$\cos 2\theta \geq \frac{1}{2}$$
You want to find the region where
$$\cos 2\theta \leq \frac{1}{2}$$
Also note that the curve ##r^2 = 6 \cos 2\theta## is not even defined for any ##\theta## such that ##\cos 2\theta < 0##. You need to take that into account as well when setting up your integral.

 

[jbunniii]

Find the area of the following region:

Inside: r² = 6 cos 2θ
Outside: r = √3

Actually there's a bit of ambiguity here. I have been assuming that this means the region whose inside curve is ##r^2 = 6 \cos 2\theta## and whose outside curve is ##r = \sqrt{3}##.

But it could also be taken to mean the region which is inside the curve ##r^2 = 6 \cos 2\theta## and outside the curve ##r = \sqrt{3}##, which is the opposite of what I assumed.

Which interpretation is correct?

 

[tolove]

Actually there's a bit of ambiguity here. I have been assuming that this means the region whose inside curve is ##r^2 = 6 \cos 2\theta## and whose outside curve is ##r = \sqrt{3}##.

But it could also be taken to mean the region which is inside the curve ##r^2 = 6 \cos 2\theta## and outside the curve ##r = \sqrt{3}##, which is the opposite of what I assumed.

Which interpretation is correct?

Ahh, I see what you're saying. The ambiguity is due to my short handing the problem.

The second of your too options.

Inside the lemniscate formed by r2 = 6cos2θ, outside the circle formed by r = sqrt 3