# Homework Help: Quick question about finding area for polar coordinates

1. Apr 30, 2012

### Bigworldjust

1. The problem statement, all variables and given/known data
Find the area of the shaded region.
r=sqrt(θ)

2. Relevant equations

A = integral from a to b 1/2r^2dθ

3. The attempt at a solution

I know how to solve the question, I just don't know what to use for a and b. I tried 0 and 2pi but I am getting the wrong answer. Here is a picture:

http://www.webassign.net/mapleplots/a/8/0c017c152b2f6bb8c6df656e3a1bc2.gif

Thank you for the help. I just need to know what parameters to use for a and b :).

2. Apr 30, 2012

### Curious3141

Which values of $\theta$ define that quadrant?

3. Apr 30, 2012

### SammyS

Staff Emeritus
If you are to find the area of the shaded region, the values of θ bounding the region look quite obvious.

You will also need to know r as a function of θ. It appears that r is directly proportional to θ. The constant of proportionality should be given in your problem.

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4. Apr 30, 2012

### andrien

can you tell the answer because it does not seem very hard

5. Apr 30, 2012

### Curious3141

Actually, his original problem statement gives $r = \sqrt{\theta}$, which makes the integration even easier.

6. Apr 30, 2012

### SammyS

Staff Emeritus
OOPS!

I missed that. r=sqrt(θ) probably would have been more recognizable with a radical or fractional exponent.

Yes, that makes things even easier!

I think we've given all the clues we should until we hear back from OP .

7. Apr 30, 2012

### Bigworldjust

Haha, wow sorry about that guys. It was late last night and I was trying to study so I was over thinking everything. Got the answer now, thanks :P.

8. Apr 30, 2012

### sharks

Yeah, i think the answer should be: $$\frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]$$

9. Apr 30, 2012

### Bigworldjust

Yeah, that was it, lol. Thanks.

10. Apr 30, 2012

### ehild

Are you sure?

ehild

11. Apr 30, 2012

### Bigworldjust

Regardless, I got the answer, lol. It was form 3pi/2 to 2pi. Sorry about asking such a dumb question guys, hah.

12. Apr 30, 2012

### sharks

Well, the integration gives:
$$A=\frac{1}{2}\left[(\ln \theta)\right] ^{\theta=2\pi}_{\theta=\frac{3\pi}{2}}=\frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]$$

Last edited: Apr 30, 2012
13. Apr 30, 2012

### skiller

Where does the $\ln \theta$ come from?

14. Apr 30, 2012

### sharks

$$A=\frac{1}{2}\int \frac{1}{\theta}\,.d\theta$$

15. Apr 30, 2012

### skiller

Are you sure? Please explain. I fail to see where you get $\frac{1}{\theta}$ from.

16. Apr 30, 2012

### sharks

Show your work and i'll tell you where you went wrong.

17. Apr 30, 2012

### skiller

Isn't the area equal to
$$\frac{1}{2}\int_\frac{3\pi}{2}^{2\pi}r^2\, d\theta$$

18. Apr 30, 2012

### sharks

You have the integrand wrong. Check the first post for the correct integrand.

19. Apr 30, 2012

### SammyS

Staff Emeritus
Yes, and $\displaystyle r=\sqrt{\theta}$ so your guess is as good as mine concerning where $\displaystyle \frac{1}{\theta}$ comes from !

20. Apr 30, 2012

### skiller

I think we can safely assume that 1/2r^2 has been misinterpreted as 1/(2r^2).

21. Apr 30, 2012

### sharks

The original integral posted by the OP is:
$$A=\int^b_a \frac{1}{2r^2}\,.d\theta$$
Since the OP has confirmed that he got the answer through this interpretation...

22. Apr 30, 2012

### Bigworldjust

Oh no that is wrong, lol. Sorry about that I didn't type it out correctly, because I didn't know how to format it on this website. What the others said as the formula were correct. Sorry about the misinterpretation.

23. Apr 30, 2012

### ehild

The area of the shaded region is $$\int_{3\pi/2}^{2\pi}{\frac{1}{2}r^2d\theta}$$
It is not the same as the integral of $\frac{1}{2r^2}$ over the shaded region.

In the second case, the integral is $$0.5\int_{\frac{3 \pi} {2}}^{2\pi} \int_0^{\sqrt{\theta}}\frac{1}{r}drd\theta$$

ehild

Last edited: Apr 30, 2012
24. Apr 30, 2012

### skiller

I don't think you should be saying this sort of thing!
No I haven't.
I don't need to; it's a commonly known integrand and even if it wasn't it's easily derived. On top of that, having double-checked the first post, it was given correctly anyway.
Indeed!
No it isn't. That's your incorrect interpretation of it.
I think the OP may have accidentally incorrectly said he came to the same answer as yours, but it was pretty obvious that they knew the integrand but was simply unsure of which upper and lower limits to take.
I think you typed it ok enough. As I said earlier, 1/2r^2 might be misinterpreted as 1/(2r^2) by some, but, if anything, the former is correct. (It reminds my of the "priority by juxtaposition" argument in the old "48÷2(9+3)" debate. (But don't get me started on that old chestnut!)
I think most (if not all) posting in this thread know this to be the case.

My guesses are one of the following (for "he", read "he or she"):

a) sharks is winding us up by picking up on an ambiguously written integrand (although I believe it is correctly written) and following through with that to the erroneous solution, in which case he really shouldn't be "confirming" a result that he knows is definitely wrong - especially in a Homework thread;

b) sharks believes he is being genuine, but only by the fact that he has assumed that the given integrand is correct (although he has incorrectly misinterpreted it), in which case he really shouldn't be "confirming" a result that he does not know necessarily to be correct - especially in a Homework thread;

c) sharks genuinely got it wrong by actually believing his integrand was correct whether he'd read it in the OP or not, in which case he really shouldn't be "confirming" results that he definitely cannot know to be true, as they are not - especially in a Homework thread.

I really don't know if it's a or b or c.

Last edited: Apr 30, 2012
25. Apr 30, 2012