Double integration and polar coordinates

[hotcommodity]

[SOLVED] Double integration and polar coordinates

Homework Statement

Find the area inside both circles r = 1, and r = 2 sin \theta by double integration in polar coordinates.

Homework Equations

None

The Attempt at a Solution

The way the problem is worded sounds a bit strange, but I believe they're asking me to find the area of a circle with radius r = 2 sin \theta minus the area or the circle with radius r = 1. I think my main problem is the way I'm setting up the double integral. When I graphed the circle with radius r = 2 sin \theta on my calculator, it didnt look anything like a circle. I graphed y = \sqrt{4sin^2 \theta - 4sin^2 \theta cos^2 \theta}. Maybe I'm graphing it wrong but here's how I set up the integral:

\int^{2 \pi}_{0} \int^{2 sin \theta}_{1} r dr d \theta

Any help is appreciated.

 

[Dick]

r=2*sin(theta) is a circle. y=r*sin(theta). So the equation becomes r=2*y/r or r^2=2y or x^2+y^2=2y. You'll need to intersect that circle with the r=1 circle before you can find the integration limits for theta. And I think what you want to do is find the area in the intersection of the two circles and use that to deduce the area of the region you want.

 

[hotcommodity]

I see, let me give that a try. Thank you for the quick response =)

 

[hotcommodity]

I understand how you arrived at x^2 + y^2 = 2y, but I'm having trouble getting y alone so that I can see what the circle looks like. This is what I've done:

x^2 + y^2 = 2y

-y^2 + 2y = x^2

- (y^2 - 2y) = x^2

I could complete the square here, but I would have to take the square root of a negative value...

 

[Dick]

You should have pushed ahead and completed the square. x^2+y^2=2y -> x^2+y^2-2y=0 -> x^2+y^2-2y+1=1. No negative problems so far...

 

[hotcommodity]

You should have pushed ahead and completed the square. x^2+y^2=2y -> x^2+y^2-2y=0 -> x^2+y^2-2y+1=1. No negative problems so far...

I'm still pretty confused about the whole thing, I know there's a key concept I must be missing. I looked at some tutorials, and they set the "r" equations equal to find the angles of intersection. When I did this for the above problem, I found that I have to integrate from \frac{\pi}{6} to \frac{5 \pi}{6}, so I'd have the integral:

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} \int^{1}_{0} r dr d \theta

I get something close to the answer, which is 1.24, but not the answer exactly.

 

[bob1182006]

your limits for r are wrong, r goes from [1,2sin\theta] or [2sin\theta,1], so just plug in some \theta \in [\frac{\pi}{6},\frac{5\pi}{6}] and determine which limits are correct and do the integration.

 

[hotcommodity]

your limits for r are wrong, r goes from [1,2sin\theta] or [2sin\theta,1], so just plug in some \theta \in [\frac{\pi}{6},\frac{5\pi}{6}] and determine which limits are correct and do the integration.

I tried that but I can't seem to get the correct answer, I'd have:

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} \int^{2sin \theta}_{1} r dr d \theta

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} \frac{r^2}{2} |^{2sin \theta}_{1} d \theta

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} 2sin^2 \theta - \frac{1}{2} d \theta

2 \int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} sin^2 \theta - \frac{1}{4} d \theta

2[ \frac{\theta}{2} - \frac{sin 2 \theta}{4} - \frac{\theta}{4}]^{\frac{5 \pi}{6}}_{\frac{\pi}{6}}

\frac{1}{2} [ \theta - sin2 \theta ]^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} = 1.91

When you graph the two circles, they overlap from r = 0 to r = 1.

 

[bob1182006]

o nvm thought the problem meant area of 2sintheta minus the r=1 portion.

So r does go from 0 to 1, what answer does the book give? it could just be wrong because that integral seems right.

 

[hotcommodity]

o nvm thought the problem meant area of 2sintheta minus the r=1 portion.

So r does go from 0 to 1, what answer does the book give? it could just be wrong because that integral seems right.

The book gives the answer \frac{4 \pi - 3 \sqrt{3}}{6}.

 

[hotcommodity]

This integral gives me the correct answer(or the books answer anyways), but it doesn't make sense to me because it's not symmetrical:

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} \int^{1}_{0} r dr d \theta + \int^{\frac{\pi}{6}}_{0} \int^{2sin \theta}_{0} r dr d \theta

I tried to add in the area cut off by the degree limits, that's why the second integral goes from zero to pi over 6. For it to be symmetrical however, I should have been able to place a 2 in front of the second integral to get the area cut off by the 5 pi over 6 limit. But the integral above gives me the same answer as the books, it just doesn't make any sense.

 

[bob1182006]

the problem is just worded wierdly ><

The area inside both r=1 AND r=2sin(theta) will just be the area inside of 2sin(theta) since 2sin(theta) is inside of r=1.

Your limits are correct for theta, but for r you want to integrate r in [0,2sin(theta)] and you should get the answer from the book, I did.

 

[hotcommodity]

If I understand you correctly, I should have:

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}}} \int^{2sin \theta}_{0} r dr d \theta

But that gives me 2.96 for an answer.

 

[bob1182006]

don't do it numerically just give the exact answer, you should get something like \frac{4\pi}{6}-\frac{\sqrt{3}}{2} which simplifies to the answer from the book.

 

[hotcommodity]

don't do it numerically just give the exact answer, you should get something like \frac{4\pi}{6}-\frac{\sqrt{3}}{2} which simplifies to the answer from the book.

Numerically the answer that the book has is 1.2284, which is nowhere near 2.96.

 

[bob1182006]

yea and that answer (4pi-3sqrt(3))/6 is 1.2284

I think you might be doing the integral wrong since I just did it again and got the same answer using the integral you have in post #13

 

[hotcommodity]

yea and that answer (4pi-3sqrt(3))/6 is 1.2284

I think you might be doing the integral wrong since I just did it again and got the same answer using the integral you have in post #13

OK, let's see...I'd have:

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}}} \int^{2sin \theta}_{0} r dr d \theta

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}}} \frac{r^2}{2} |^{2sin \theta}_{0} d \theta

2 \int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}}} sin^2 \theta d \theta

2 \left[\frac{\theta}{2} - \frac{sin2 \theta}{4} \right]^{\frac{5 \pi}{6}}_{\frac{\pi}{6}}} = 2[ 1.53 - 0.0453] = 2.96

If I'm evaluating the integral wrong, I have no idea how.

Edit: Also, the bounds for r here don't really represent the area that's being overlapped. If you have a graphing calculator, put the graph in polar coordinates and I think you'll see what I mean. The area that overlaps is from r = 0 to r =1.

 

[bob1182006]

>< missed a - sign.
did the book really say 4pi/6 - ? not +? I am getting + and doing it using the symmetry of the circle:
2\int_\frac{\pi}{6}^\frac{\pi}{2}d\theta \int_0^{2sin\theta}r dr

since pi/6 to pi/2 is the same area as pi/2 to 5pi/6.

and the final answer I get is \frac{4\pi+3\sqrt{3}}{6}

I think the book just had a typo at this problem..

 

[hotcommodity]

It's definitely a minus sign. I just reworked the integral you posted with the difference in radian limits, and I still get 2.96. I have no clue how you're getting your answer =/

 

[bob1182006]

2.96=what I got just mine still has the pi in it.

well it's really easy to say - instead of +, if it wasn't a typo then probably the problem is just worded incorrectly and they wanted something else...

 

[hotcommodity]

Could be, I'll find out tomorrow. Thanks for trying to help me out =)

 

[Dick]

This integral gives me the correct answer(or the books answer anyways), but it doesn't make sense to me because it's not symmetrical:

\int^{\frac{5 \pi}{6}}_{\frac{\pi}{6}} \int^{1}_{0} r dr d \theta + \int^{\frac{\pi}{6}}_{0} \int^{2sin \theta}_{0} r dr d \theta

I tried to add in the area cut off by the degree limits, that's why the second integral goes from zero to pi over 6. For it to be symmetrical however, I should have been able to place a 2 in front of the second integral to get the area cut off by the 5 pi over 6 limit. But the integral above gives me the same answer as the books, it just doesn't make any sense.

That's it. Yes, double the second integral or just write it again changing the limits to 5pi/6 to pi. There are two regions where the r=2*sin(theta) circle are inside the r=1 circle and one where it is outside.

 

[hotcommodity]

Oh you're right! haha. Finally it's demystified :) Thank you both for your help.