Question

find the area enclosed by one loop of the four-leaved rose: r=8sin6theta

I know that to obtain the limits you set r=0, but so far have only practiced with r=4+3sintheta, not with coefficients.

Answer 1

Setting r=0? Hmm yah I think that will work nicely.

\[\large r=8\sin(6\theta) \qquad \rightarrow \qquad 0=8\sin(6\theta)\]Dividing both sides by 8 gives us,\[\large 0=\sin(6\theta)\]

Let's think about the first time when the sine of an angle produces a value of 0.
That's at 0 right?\[\large 6\theta=0\]

When would be the next time sine of an angle produces 0?
\[\large 6\theta=\pi\]
Right?
From here, just solve for theta to get your limits of integration.
If you're confused, let me know.

Answer 2

ok, i understand how you get pi. but how would you figure out the other limit, since we already know what you get when you solve for theta