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## Homework Statement

I am after finding the centroid of the remaining area (hatched) when a circle is cut by a line. I made a diagram in CAD that demonstrates the problem.

The idea is that, starting from the bottom of the circle, a cut is taken leaving a remaining shape whose area and centroid can be calculated. [itex]dc[/itex] denotes the height from the bottom of the circle to the center of mass and [itex]c[/itex] is the height from the bottom of the circle where the cut line is.

I believe the problem I am having is to do with the equation of the circle being written in polar coordinates and the equation of the line is in cartesian coordinates.

## Homework Equations

The mass of a lamina is per the following definition:

[tex]M = ρ\int_{a}^{b} f(x)-g(x) dx[/tex]

Where I am taking:

$$f(x) = \sqrt{r^2-x^2}$$ and,

$$g(x) = c$$

I am ignoring the density [itex]ρ[/itex] in the mass equation above because it will cancel out when I divide using moments to solve for the center of mass:

[tex]M_x = ρ\int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) dx[/tex]

Thus the center of mass of a lamina is:

[tex]\bar{x}=\frac{M_x}{M}=\frac{\int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) dx}{\int_{a}^{b} f(x)-g(x) dx}=\frac{1}{A}\int_{a}^{b} \frac{1}{2}([f(x)]^2-[g(x)]^2) dx[/tex]

## The Attempt at a Solution

My first step is to use the same coordinate system for both the line at distance [itex]c[/itex] and the circle.

The equation of the circle is thus:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

rewritten in terms of y:

[tex]f(x) = \sqrt{r^2-(x-h)^2}+k[/tex]

Where [itex](h,k)[/itex] represents in the cartesian [itex](x,y)[/itex] coordinates that define the center of the circle. Now is makes more sense that I also use:

[tex]g(x) = c[/tex]

Next step is solving for the bounds on the above intergrals [itex]a,b[/itex].

[tex]x = \sqrt{r^2-(c-k)^2}+h[/tex]

Thus:

[tex] b = +\sqrt{r^2-(c-k)^2}+h[/tex]

[tex] a = -\sqrt{r^2-(c-k)^2}+h[/tex]

Mass integral with constant density:

[tex]M = ρ\int_{-\sqrt{r^2-(c-k)^2}+h}^{+\sqrt{r^2-(c-k)^2}+h} (\sqrt{r^2-(x-h)^2}+k-c) dx[/tex]

I want to hold here for some further advice because the resulting expression is quite large. Does anyone have any comments on my working so far? Is there an easier approach? Perhaps remaining in polar coordinates?