# Check this question to see if it is right.

1. Oct 5, 2007

### yaho8888

1. The problem statement, all variables and given/known data
Find area between the polar graph$$r= \theta$$ and $$r=2\theta$$
from $$0 \leq \theta \leq 2\pi$$.

2. Relevant equations
$$Area = \frac{1}{2} \int r^2 \theta$$

3. The solution

$$Area = \frac{1}{2} \int_0^2 (2 \theta)^2 d\theta - \frac{1}{2} \int_0^2 \theta^2 d\theta$$

final answer $$4\pi^3$$ or 124.025

check please as soon as possible, only have to say right or wrong!

Last edited: Oct 5, 2007
2. Oct 5, 2007

### mlazos

I put the data to mathematica by using the integrals you wrote and i got 1.0472. I dont understand how you derived the integral and what exactly you ask. Please re write the question so we can understand better what you want.

3. Oct 5, 2007

### bob1182006

looks wrong,

for a polar area your integral should be:
$$\frac{1}{2}\int_{a}^{b}r^2 d\theta$$

also your a/b seem wrong, it should be 0 and 2pi but you have 0 and 2

4. Oct 5, 2007

### yaho8888

sorry i type some error, I edit the question above!
check again!

5. Oct 5, 2007

### yaho8888

is from 0 to 2pi for the integrals

6. Oct 5, 2007

### mlazos

According to your integrals and to your latest corrections the value you get is correct. I double checked with mathematica and the answer is ok.

7. Oct 5, 2007

### bob1182006

yep that answer seems correct now

8. Oct 5, 2007

Thanks!