Hi I am having trouble trying to figure out how to find the area of the region enclosed by one loop of the curve r=10sin(2theta)?

Question

anonymous · Answer

$$\int\limits_{0}^{360} \int\limits_{0}^{10\sin(2\theta)} r*dr * d \theta $$

because r varies from 0 to 10 sin(2theata).

r dr dtheata is the standard way of doing a double int in polar form.

anonymous · Answer

wouldn't i have to square the r in the formula?

anonymous · Answer

I'm a little fuzzy on this, but I don't believe so. Because really you're just taking the integral of 1 * r dr dtheata,

The r becomes square when you take the first integral.  When you apply the bounds 10sin(2theata) drops in, the zero bound goes away.

anonymous · Answer

oh okokok thanks