# Solve for the center of mass of the 1st-Quadrant portion of the unit circle centered at the origin of radius 1.

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The coordinates of the center of mass of a plate are determined by the formulas

`x_c=1/A int^b_axf(x)dx` and `y_c=1/A int^b_a 1/2 (f(x))^2dx` , where A is the area of the plate. In case of the first-quadrant portion of the unit circle, the area is a quarter of the area of the circle with radius 1, `pi*1^2 = pi` . So `A = pi/4` .

The formula f(x) of the first-quadrant portion of the unit circle is

`y = sqrt(1 - x^2)` with lower limit a = 0 and upper limit b = 1 (see the graph below.)

Consider the integral for the x-coordinate of the center of mass:

`int^1_0 xsqrt(1-x^2)dx` . Use substitution to evaluate this integral:

`u = 1 - x^2` ; u(0) = 1 and u(1) = 0

`du = -2xdx` , from where `xdx = (du)/-2`

Putting this into the integral, obtain

`int^0_1 sqrt(u) (du)/-2=1/2 int^1_0 u^(1/2)du = 1/2*2/3*u^(3/2) |^1_0=1/3`

Dividing this by the area of the quarter-circle, we get

`x_c = (1/3)/(pi/4) = 4/(3pi)`

Now let's evaluate the intergral for the y-coordinate of the center of mass. Because the quarter-circle is symmetric around the line y = x, we can expect `y_c` to be equal to `x_c` .

`int^1_0 1/2 (sqrt(1-x^2))^2 dx = 1/2 int^1_0 (1 - x^2)dx = `

`1/2 (x - x^3/3) |^1_0 = 1/2(1 - 1/3) = 1/3`

Dividing this by `A = pi/4` we get `y_c = 4/(3pi)` , as expected.

**The coordinates of the center of mass of the first-quadrant unit circle centered at the origin are**

`x_c = y_c = 4/(3pi)`