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Find the area of the region inside both polar graphs

  • Thread starter Pi Face
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  • #1
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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.
 

Answers and Replies

  • #2
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Set the two r values equal to each other -- you can't set two equations equal to each other -- which gives you 1 + sin([itex]\theta[/itex]) = 5sin([itex]\theta[/itex]).

The graph of r = 5sin([itex]\theta[/itex]) 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.
 
Last edited:
  • #3
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but how do I find the numerical values of these two points?
 
  • #4
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Solve the equation 1 + sin([itex]\theta[/itex]) = 5sin([itex]\theta[/itex]). That will give you one of the values of [itex]\theta[/itex]. And because sin([itex]\pi - \theta[/itex]) = sin([itex]\theta[/itex]), that will give you the other one.
 
  • #5
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is there a way to solve it manually? my ti-89 gives me 6.28....(2pi) and 2.8889, which im not sure what that converts to in radians.
 
  • #6
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which on second thought, doesnt make sense.

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

huh?
 
  • #7
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is there a way to solve it manually? my ti-89 gives me 6.28....(2pi) and 2.8889, which im 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([itex]\theta[/itex]) = 5sin([itex]\theta[/itex])? Show me your steps. No calculator.
 
  • #8
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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 gotta go over my trig.
 
  • #9
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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.
 
  • #10
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Got it. The actual finding the area part was easy, just blanked out at the sinx=1/4 part. thanks for you help
 

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