Question

r=5cos(5theta) I need to find the area enclosed by one leaf. I know that this is a double integral problem, and I know that r goes from 0 to 5cos(5theta), but I don't know how to get the limits of integration for the angle...

Answer 1

why a double integral?

Answer 2

well I suppose it's the only way I know how to do it right now, but this is how the test question is gonna be. plus I don't know how to do regular integrals in polar anyway

Answer 3

http://www.wolframalpha.com/input/?i=r%3D5+cos%285+theta%29

Answer 4

so the only way to do it is to have a picture of it?

Answer 5

just take
\[25\int_0^{\frac{\pi}{10}} \cos(5\theta) d\theta\]

Answer 6

in general it is
\[\int_a^b \frac{1}{2}r^2 d\theta\]

Answer 7

i was integrating so that you go from 0 to pi/2, which gives you half the petal, then double. that takes care of the 1/2

Answer 8

and of course pull the 5^2=25 out of the integral

Answer 9

oh crap i had a typo. it should be cosine squared. sorry

Answer 10

\[25\int_0^{\frac{\pi}{10}} \cos^2(5\theta) d\theta\]

Answer 11

\[\int\limits_{\theta=a}^{\theta=b}\int\limits_{r=c}^{r=d}rdrd \theta\]

Answer 12

have fun. but you do not need a double integral for this, that is for sure.

Answer 13

is what it's supposed to look like, but I can't figure out the theta's limits. and 0 to pi/2 gives you 1.5 petals of the 5cos(5theta)

Answer 14

k thanks

Answer 15

that is why i am integrating from 0 to pi/10

Answer 16

you want
\[5\theta=\frac{\pi}{2}\] ,making
\[\theta=\frac{\pi}{10}\]

Answer 17

so integrate from 0 to pi/10, not from 0 to pi/2

Answer 18

why do you set the 5theta equal to pi/2?