hello everyone can somebody help me with this question?
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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
in the answer to this question they got "pi/6" where the R disappear ??
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
There is a problem with the solutions how do you integrate the INTEGRAL(cos^2(theta)*sin^2(theta))
Last edited by sbsite; Aug 10th 2007 at 03:48 PM.
Rewrite the it as $\displaystyle cos^{2}{\theta}sin^{2}{\theta}=\frac{1}{8}(1-cos{4\theta})$
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
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