# 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.

#### Attached Files:

• ###### 0c017c152b2f6bb8c6df656e3a1bc2.gif
File size:
3.3 KB
Views:
5,413
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

### DryRun

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

### DryRun

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

### DryRun

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

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

### DryRun

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

### DryRun

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).