Question

Find the area that the graph encloses:

r=1+2sin (6theta)

I will provide a graph your convenience.

Answer 1

I know how to evaluate integrals like this using polar curves
\[\int\limits_{a}^{b}(1/2)r^2d \theta\]
My trouble lies with finding the bounds of one loop of the graph. Also I assume that this will be the addition of two integrals, one for the small loop and one for the big one (and obviously multiplied by the number of respective loops)

Answer 2

hey this is a stupid question
so when ever you want to graph a polar equation you use:
\[\int\limits\limits_{a}^{b}(1/2)r^2d \theta \]?

Answer 3

i mean find area not graph

Answer 4

correct, where r is a function of theta. In this case 1+2sin (6theta)

Answer 5

so we want to find the area enclosed by r
don't we just want to find when r is 0?

Answer 6

for the bounds

Answer 7

i need an example
let me see if i can find one
or if you know a good site let me know

Answer 8

Yes, I need the bounds for both the loops or when r=0. I calculated r for theta at pi/6 intervals from 0 to pi and I got that every r was = 1 every time. However I just realized I skipped a few so im gonna double check real quick. And ill try to find some online examples

Answer 9

I managed to graph this on geogebra. The period is 2pi and the integral (over the full period) should be

\(\frac{1}{2}\int_0^{2\pi}(1+2\sin(6\theta))^2 d\theta\)

The geogebra plot of it is pretty sweet. Let me know if you'd like to see how it's traced out over time.

Answer 10

I realized that I was making it more difficult than it had to be by finding the area of one loop and multiplying that by the number of loops. When really I just needed to find the bounds of the entire graph, as you suggested mathteacher\[0\le \theta \le2\pi\]
and I got the correct answer being 3pi.

Answer 11

awesome