Find the area of the region inside both polar graphs

[Pi Face]

Homework Statement

r1= 1+sin(theta)
r2= 5sin(theta)

Homework Equations

see above?

The Attempt at a Solution

totally stumped. usually i would set the two curves equal to each other, but i have no idea how to do that. using my ti-89's solve function just gives me a weird answer using arcsin as well.

 

[Mark44]

Set the two r values equal to each other -- you can't set two equations equal to each other -- which gives you 1 + sin($\theta$) = 5sin($\theta$).

The graph of r = 5sin($\theta$) is a circle of radius 2.5, whose center is at (0, 2.5) in rectangular coordinates. The other curve is a cardioid, a kind of heart-shaped curve. These curves intersect at two points.

 

[Pi Face]

but how do I find the numerical values of these two points?

 

[Mark44]

Solve the equation 1 + sin($\theta$) = 5sin($\theta$). That will give you one of the values of $\theta$. And because sin($\pi - \theta$) = sin($\theta$), that will give you the other one.

 

[Pi Face]

is there a way to solve it manually? my ti-89 gives me 6.28...(2pi) and 2.8889, which I am not sure what that converts to in radians.

 

[Pi Face]

which on second thought, doesn't make sense.

1+sin(2pi)=5sin(2pi)
1+0=0
1=0

huh?

 

[Mark44]

is there a way to solve it manually? my ti-89 gives me 6.28...(2pi) and 2.8889, which I am not sure what that converts to in radians.

Absolutely, there's a way to solve it manually. Nothing I have done required the use of a calculator of any kind.

2.8889 is in radians, but both answers you show are WAY off. I really think you should put your calculator away.

What exactly did you do to solve 1 + sin($\theta$) = 5sin($\theta$)? Show me your steps. No calculator.

 

[Pi Face]

Just woke up. A little groggy but I'll give it a shot.
theta=x out of laziness

1+sin(x)=5sin(x)
1=4sin(x)
sin(x)=1/4
...now what?
no where out of the "standard" points on the unit circle (pi/3,pi/2,pi/4, etc) does the y value equal 1/4

unless I can use the half/double angle formula somehow?

I REALLY got to go over my trig.

 

[Mark44]

OK, good so far.
sin(x) = 1/4 ==> x = arcsin(1/4) = sin^-1(1/4). Now you can use a calculator to get an approximate value for x, which is about 14.5 degrees, or about .253 radians. The exact value is The other angle is pi - x, or about 165.5 degrees.

Now, to find the area that is inside both curves, you should probably set this up as a polar integral. You can make life slightly easier by using the symmetry here, noting that both curves are symmetric about the y axis. Draw a graph of the two curves so you can figure out what your integrand needs to be. Also, because the boundary changes at the points of intersection, you'll need two integrals.

 

[Pi Face]

Got it. The actual finding the area part was easy, just blanked out at the sinx=1/4 part. thanks for you help