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Find area of inner curve polar equation

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

Sketch $\displaystyle r=3-5 \cos {theta}$ 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.)

Attachment 30044

I'm supposed to find the area of the curve from $\displaystyle r=-2$ to $\displaystyle 0$ and when $\displaystyle \theta$ goes from ?? This is where I get stuck. To find where $\displaystyle r=0$, that is, where the curve is at O and a tangent there, $\displaystyle 3=5\cos(\theta)$ I need two values for $\displaystyle \theta$ 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 $\displaystyle ~53.13010235$ which converting to radians I take to be $\displaystyle \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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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 $\displaystyle \frac{3\pi}{2}$ 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.