Area in cardioid and outside circle - Using Double Integral

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Area in cardioid and outside circle -- Using Double Integral

Homework Statement



Find the area inside of the cardioid given by r = 1 + cos[itex]\theta[/itex] and outside of the circle given by r = 3cos[itex]\theta[/itex].

Homework Equations


[itex]\int[/itex][itex]\int[/itex]f(x,y)dA = [itex]\int[/itex][itex]\int[/itex]f(r,[itex]\theta[/itex])rdrd[itex]\theta[/itex]

not really relevant, as it wasnt given in rectangular coordinates...oh well =s

The Attempt at a Solution



Find theta of intersection of the two lines (above x axis):

1 + cos[itex]\theta[/itex] = 3cos[itex]\theta[/itex] => 2cos[itex]\theta[/itex]= 1 => [itex]\theta[/itex] = cos-1(1/2)= [itex]\pi[/itex]/3

Looking at the graph of the situation, there are two regions. One region with the cardioid ontop of the circle, and one with only the cardioid. This first region (the region with the circle underneath the cardioid) goes from the intersection at [itex]\pi[/itex]/3, to [itex]\pi[/itex]/2.

Because the cardioid is on top, and the circle is on the bottom (for this region), we have:
[itex]\int^{\pi/2}_{\pi/3}[/itex][itex]\int^{1+cos\theta}_{3cos\theta}[/itex]drd[itex]\theta[/itex]

For the next region (the other side of the y-axis), the radius is given solely by the function for the cardioid, and the angle is from [itex]\pi[/itex]/2 to [itex]\pi[/itex]. The total area (via symmetry) is given by the sum of these two integrals multiplied by two.

A = 2 ( [itex]\int^{\pi/2}_{\pi/3}[/itex][itex]\int^{1+cos\theta}_{3cos\theta}[/itex]drd[itex]\theta[/itex] + [itex]\int^{\pi}_{\pi/2}[/itex][itex]\int^{3cos\theta}_{0}[/itex]drd[itex]\theta[/itex])

However, this integral produces the wrong answer. I think that the problem is with my limits for r on the first integral above ([itex]\int^{1+cos\theta}_{3cos\theta}[/itex]), however I am not quite sure what the problem is..


Can anyone give me a hint or a hand (though from my experience homework helpers on physics forums arent into hand holding as far as problem solving goes.. =])

the reason I chose those limits for the radius stems from the fact that the circle is on the bottom and the cardioid is on top [from the circle, to the cardioid --- from 3cos(t) to 1 + cos(t)]


thanks in advance
 

Answers and Replies

  • #2
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Do you know that the area in polar coordinates can be represented by:

[itex]\int_a^b {r^{2}d\theta}[/itex]

where a and b are the values of theta you want the area between. I think this approach is clearer.
 
  • #3
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Yes but this is for a multivariable calc class and this question asks for this to be done with a double integral.
I thought I said this. If not, sorry.
 
  • #4
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It seems that your integral is missing an r, i.e. it should be an integral of rdrdθ
 
  • #5
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oh duh i am so dumb... dA = rdrd[itex]\theta[/itex] ><
 
  • #6
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would that matter though? I didnt convert it from Cartesian coordinates..
 
  • #7
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Yes, it would, since the area of the "infinitesimal" in polar coordinates is always rdrdθ, not drdθ (which you can see from a diagram). It doesn't matter if you convert or not, when you perform such integrations, you have to include this factor.
 
  • #8
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A = [itex]\int\int[/itex]dA = [itex]\int\int rdrd\theta[/itex]

just saw a theorem in my book. Thanks for your help =s
 

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