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Find area of inner curve polar equation

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

Sketch 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.)

Attachment 30044

I'm supposed to find the area of the curve from to and when goes from ?? This is where I get stuck. To find where , that is, where the curve is at O and a tangent there, I need two values for 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 which converting to radians I take to be . 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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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.

Attachment 30046

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 appear? I'm stumped. Any help is greatly appreciated!!

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 x_{1} and x_{2} in (0,2pi) with cos(x)=v. The first is x_{1}=acos(v) and then x_{2}=2pi-x_{1}=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.