Question

I can't seem to get a value for the area between these two polar curves: 2+sin(theta) and 3sin(theta)

Area = .5 * Integral of (r^2 * dtheta)

Answer 1

I know its larger area - smaller area, but I think my bounds of pi/2 and 5pi/2 are incorrect

Answer 2

First things first, I'm absolutely unsure about what follows, but I think your bounds are alright actually. These are the points where both curves have the same value, 3. Now for the area, I integrated both of these equations and got this :
2theta - cos(theta) and -3cos(theta)
for the bounds, you'll get 5pi for (5pi/2) and pi for (pi/2), which would mean that your area is 4pi?
Again, not sure, but that's how I'd solve this

Answer 3

not sure if this is what you did, but I learned you had to subtract the larger area from the smaller area, so [2+sinx]^2 - [3sinx]^2 which gives me .5 * Integral of [4+4sinx-8sin^2x]

Answer 4

when I use the bounds I keep getting 0 for the area which is obviously wrong.

Answer 5

hmm, why do you square the functions? I don't think I ever did that

Answer 6

oh, it's because of the polar curves, is it? Well, I think I'm a bit over my head in this one, sorry

Answer 7

that's ok thanks for trying

Answer 8

I think that you need to do something special with your bounds because of the fact that both curves only intersect once over a period :-S

Answer 9

they do intersect but the 1st equation is always larger. How did you settle on your bounds? I believe it should be 0 to 2pi

Answer 10

I set the two equations equal to each other and got sin(theta) = 1, which is pi/2 and 5pi/2

Answer 11

\[\frac{5 \pi}{2}=\frac{pi}{2} +2 \pi \]
meaning that this point is the same point only on the second revolution of the sin period

Answer 12

http://fooplot.com/plot/pnttal57f7

Answer 13

yes so don't I need two bounds?

Answer 14

0 and 2pi have a difference of 2pi as well

Answer 15

your bounds would be 0 and 2pi. this is how you express a complete revolution in the polar coordinate system.
0 and 2pi aren't the same, but if you plug those values into a function that has a period of 2pi, then you will get the same value out of the function

Answer 16

sin and cos functions are two examples of functions that have this period