# What is the equation for the area that is inside of r=cos(x)+1 and outside of r=cos(x)?

In class, my professor said that:

1/2integral from 0 to 2pi

((cosx+1)^2-cos^2x)dx is not the correct formula for this equation. However, she gave the area formula as:

integral from b to a, 1/2(f^2(x)-g^2(x)).

Thank you for any clarification.

### 1 Answer

- RealProLv 710 months agoFavorite Answer
What I would like to know is why your teacher uses "x" for angle. I've never seen anyone do it and it only leads to confusion.

It doesn't look like you tried graphing it. It's very special when you can visually follow what's happening.

The 2nd curve r=cos(θ) is completed already for 0 < θ < pi. Tracing out the cosθ counterclockwise quadrant by quadrant will make you see why.

0 < θ < 2pi draws the curve twice, "over itself", and so integrating from 0 to 2pi gives twice the area bounded by the curve which is not what you want to be subtracting.

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We can calculate the area of the first curve using the regular formula

1/2 * [int from 0 to 2pi] ((cosθ + 1)^2)dθ

And then (since the 2nd curve is entirely inside the 1st one) we subtract the area of the 2nd,

1/2 * [int from 0 to pi] ((cosθ)^2)dθ

Note the different limits.

Often in polar coordinates for calculating area you need more than one integral due to the mismatch between the angles that are needed to "finish" the graph and hence the bounds of the integrals.

I didn't have a symbol for theta, so I just used X and I did try graphing it, I know what it looks like, but that didn't make it any clearer. I appreciate your answer, but you don't need to tear someone down because they're not understanding it.