Check this question to see if it is right.

[yaho8888]

Homework Statement

Find area between the polar graph$$r= \theta$$ and $$r=2\theta$$
from $$0 \leq \theta \leq 2\pi$$.

Homework 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!

 

[mlazos]

I put the data to mathematica by using the integrals you wrote and i got 1.0472. I don't understand how you derived the integral and what exactly you ask. Please re write the question so we can understand better what you want.

 

[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

 

[yaho8888]

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

 

[yaho8888]

is from 0 to 2pi for the integrals

 

[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.

 

[bob1182006]

yep that answer seems correct now

 

[yaho8888]

Thanks!