Question

Find the area inside of r = 3 + 3sin(theta)

Answer 1

Umm ok I'm trying to remember how to do these... Let's see..

So the formula for area in polar is:\[\large \int\limits_a^b \frac{1}{2}r^2\;d \theta\]Right?

Soooooo, we're given our function r.
Were you able to find the limits of integration?
We could graph it, that might help.

Answer 2

It should be 27pi/2 and I'm trying to solve it using a double integral.

So it will be:
\[\int\limits_{a}^{b}\int\limits_{f(\theta)}^{g(\theta)}rdrd \theta\]

because the volume element, dA, is r dr dtheta, in polar coordinates

Answer 3

I'm just having trouble finding the limits

Answer 4

Here is the graph

Answer 5

EDIT: earlier I said 'volume element' I meant area element

Answer 6

I think what we want to do is.....
We want the area to sweep around from that origin point.

So we would START from an angle of \(\large \theta=-\pi/2\) and sweep around and END at \(\large \theta=3\pi/2\)

Answer 7

You can see that the distance we cover is 2pi.
But I think if we integrate from 0 to 2pi, we might run into trouble.

I'm not sure though :o I'd have to try it both ways to see...

Answer 8

then the limits on r would be 0 to 3 + 3sin(theta) ?

Answer 9

Our integral is being taken with respect to theta, so the limits of integration are with respect to theta.
\[\huge \int\limits\limits_{\theta=a}^b \frac{1}{2}r^2\;d \theta\]

Where \(\large a=-\pi/2\) and \(\large b=3\pi/2\) in this case.

Maybe I misunderstood what you were saying though.
Limits on r? Hmm.
We don't have anything to subtract from r.
But yes as you said we're going from 0 to our function r.

If you wanted to, you could write your function as [3+3sin(theta)-0]. But blah.

Answer 10

\[\huge \int\limits\limits\limits_{\theta=-\pi/2}^{3\pi/2} \frac{1}{2}(3+3\sin \theta)^2\;d \theta\]

Answer 11

That results in 3pi.

Also, I know it can be done using double integrals.

Answer 12

3pi? ok maybe i made a mistake somewhere, gimme a sec :)

Answer 13

But I'm trying to use this method: http://tutorial.math.lamar.edu/Classes/CalcIII/DIPolarCoords.aspx

Answer 14

I understand how it works but I don't know how to setup the limits in this example.

Answer 15

Ex 2 from that site is very similar. But when integrating from r=0 to r=3+3sin(theta), I don't know what the theta limits should be.

Answer 16

Where are you getting 3pi from?
That sounds way off...

Answer 17

I'm doing all the integration in matlab. But I've solved it now.

Answer 18

I've been literally mistyping the 3 + 3sin(theta) limit EVERYTIME as 3sin(theta) XD!!!

Answer 19

Oh lololol

Answer 20

If you were confused about the way they got the limits of integration, we can do the same thing as in the example ~ Find points of intersection.

When does r=0 `intersect` r=3+3sin(theta) ?

If you've got this all figured out already then no problem. Heh.

Answer 21

So the solution is:
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{3 + 3\sin(\theta)}r dr d \theta\]

Thanks for your time though!

You've kept me looking at this problem long enough to notice my mistake, and I appreciate this!

Answer 22

I think I need some sleep...

Thanks again!

Answer 23

Mmmm that 0 and 2pi don't look correct :[
I think those would give you a problem if you use them for your limits on theta...

Answer 24

But whatever, go to sleep if you're tired! lol