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Math Help - Find area of inner curve polar equation

  1. #1
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    Find area of inner curve polar equation

    Hi everyone again! My question is as follows in full:

    Sketch  r=3-5 \cos {theta} and find the area of the inner curve.

    Ok, this is a cardioid [according to my book], and my graphing by sketch looks like (couldn't scan so this is an actual graph courtesy of desmos.)

    Find area of inner curve polar equation-r-3-5costheta.jpg

    I'm supposed to find the area of the curve from r=-2 to 0 and when \theta goes from ?? This is where I get stuck. To find where r=0, that is, where the curve is at O and a tangent there, 3=5\cos(\theta) I need two values for \theta but I can only find one, which logically makes two because both the plus and minus of the absolute value of that value should apply.

    The value I found is ~53.13010235 which converting to radians I take to be \pm 0.295167235 \pi . If I apply this number backwards from [tex]\pi[\tex], I can't seem to get the correct value. I seem to be getting the whole sector. Some help here please?

    Thanks in advance!
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  2. #2
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    Re: Find area of inner curve polar equation

    Hi,
    The attachment shows a more detailed sketch of your curve. Follow the arrows to see how the curve is traced out from 0 to 2pi. Left to you is the exact determination of the bounding curves for your region and also the actual computation of the integrals -- the answers are given but, after all, this is your problem.

    Find area of inner curve polar equation-mhfcalc27.png
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  3. #3
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    Re: Find area of inner curve polar equation

    The part of the problem that I was having trouble with is how to find the boundary of the inner curve, that is, the arcsin/arccos part.

    I now understand the arcsin (3/5) and arccos(3/5), but how in the world does \frac{3\pi}{2} appear? I'm stumped. Any help is greatly appreciated!!
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  4. #4
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    Re: Find area of inner curve polar equation

    Hi again,
    As you originally pointed out, you want the 2 theta values where r = 0. In hindsight, I probably shouldn't have used 3pi/2+asin(3/5) for the 2nd theta value. (I wanted to show on the graph that this 2nd value is greater than 3pi/2.)

    For any value v in (-1,1), there are two x values x1 and x2 in (0,2pi) with cos(x)=v. The first is x1=acos(v) and then x2=2pi-x1=2pi-acos(v). So the 2nd theta value where r = 0 is 2pi-acos(3/5).

    Now for any x in [-1,1], acos(x)+asin(x)=pi/2. (The easy way to see this is that the derivative of acos(x)+asin(x) is 0, and so this must be a constant. Plugin some x value, say x=0 to see what the constant is.) So 2pi-acos(3/5)=2pi-(pi/2-asin(3/5))=3pi/2+asin(3/5).

    I hope this clears up any confusion on your part.
    Last edited by johng; January 20th 2014 at 09:15 AM.
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