Find the area of the region enclosed by one loop of the curve. r=2sin5θ

Question

anonymous · Answer

A loop looks like goes from 0 to pi/5 so integrate
1/2 (3 sin 5 theta) over that interval should give you the area.

anonymous · Answer

can you help me solve that step by step?

anonymous · Answer

Actually, I made a couple booboos there now I am looking at it again

Should be 1/2 (2 sin 5 theta)^2  ie just 2 sin^2 (5 theta)

anonymous · Answer

okay i got that far.

anonymous · Answer

Area of one loop = pi/5 ?

anonymous · Answer

Maybe you can use double angle formula to get an easier integral....

anonymous · Answer

I haven't actually worked it out myself....

anonymous · Answer

sin^2 5theta = (half theta - sin 10 theta)/20  maybe...

anonymous · Answer

sin switches to cos in the double angle formula

experimentx · Answer

$$ 1/2 \int_{\theta_0}^{\theta_1} r^2 d\theta$$

anonymous · Answer

is the answer pi/5 ?`

experimentx · Answer

anonymous · Answer

one leaf/petal of the flower shape

experimentx · Answer

$$ \theta_0 = 0 $$ $$ \theta_1 = \pi/5 $$

anonymous · Answer

yes those are the theta bounds for the integral

experimentx · Answer

anonymous · Answer

okay thank you!!!

experimentx · Answer

yw