1. ## double integrals

hello everyone
can somebody help me with this question?

2. Originally Posted by sbsite
hello everyone
can somebody help me with this question?
Introduce polar variables $\displaystyle (r,\theta)$ such that $\displaystyle x=r \cos(\theta),~y=r \sin(\theta)$, then the integaral becomes:

$\displaystyle \int_{\theta=0}^{2\pi} \int_{r=0}^R r^4 \cos^2(\theta) \sin^2(\theta) ~r ~d\theta dr$

RonL

3. ## Thank you

in the answer to this question they got "pi/6"
where the R disappear ??

4. Originally Posted by sbsite

in the answer to this question they got "pi/6"
where the R disappear ??
I get $\displaystyle \frac{\pi R^6}{24}$

-Dan

5. ## Thank you

There is a problem with the solutions
how do you integrate the INTEGRAL(cos^2(theta)*sin^2(theta))

6. Rewrite the it as $\displaystyle cos^{2}{\theta}sin^{2}{\theta}=\frac{1}{8}(1-cos{4\theta})$

7. Originally Posted by sbsite
There is a problem with the solutions
how do you integrate the INTEGRAL(cos^2(theta)*sin^2(theta))
Or, alternately and a bit more complicated, you could do
$\displaystyle (sin(\theta)cos(\theta))^2 = \left ( \frac{1}{2} sin(2\theta) \right ) ^2$

-Dan