The Centroid

We now have expressions for M_x and M_y. To get the coordinates of the centroid we use the fact that these moments are equal to the ones you would get by concentrating the whole mass at the centre of gravity.

Let the centre of gravity be at ($\bar{x}$,$\bar{y}$), and let the area of the region be A.

The distance of the centre of gravity from the x-axis is $\bar{y}$, and so the moment about the x-axis is A$\bar{y}$. Similarly that about the y-axis is A$\bar{x}$. Therefore we have the formulas

A$$\bar{x}$$ = M_y ,        A$$\bar{y}$$ = M_x

We know how to calculate A, M_x and M_y, and so using these formulas we can calculate $\bar{x}$ and $\bar{y}$.

Example 8..6   Find the centroid of the region bounded by y = $\sqrt{x}$, the x-axis and the line x = 1.

Solution Calculate A, M_x, M_y and then put them into the formulas.

A = $$\int_{0}^{1}$$$$\sqrt{x}$$ dx = $$\left[\vphantom{\frac{2}{3}x^{3/2}}\right.$$$${\frac{{2}}{{3}}}$$x^3/2$$\left.\vphantom{\frac{2}{3}x^{3/2}}\right]_{0}^{1}$$ = $${\frac{{2}}{{3}}}$$

For the two moments we can use the simplified formulas, since the lower boundary is the x-axis.

M_x = $$\int_{0}^{1}$$$${\frac{{1}}{{2}}}$$($$\sqrt{x}$$)² dx = $$\int_{0}^{1}$$$${\frac{{1}}{{2}}}$$x dx = $$\left[\vphantom{\frac{1}{4}x^2}\right.$$$${\frac{{1}}{{4}}}$$x²$$\left.\vphantom{\frac{1}{4}x^2}\right]_{0}^{1}$$ = $${\frac{{1}}{{4}}}$$

M_y = $$\int_{0}^{1}$$x$$\sqrt{x}$$ dx = $$\int_{0}^{1}$$x^3/2 dx = $$\left[\vphantom{\frac{2}{5}x^{5/2}}\right.$$$${\frac{{2}}{{5}}}$$x^5/2$$\left.\vphantom{\frac{2}{5}x^{5/2}}\right]_{0}^{1}$$ = $${\frac{{2}}{{5}}}$$

Therefore

$$\bar{x}$$ = $${\frac{{M_y}}{{A}}}$$ = $${\frac{{2}}{{5}}}$$$${\frac{{3}}{{2}}}$$ = $${\frac{{3}}{{5}}}$$

and

$$\bar{y}$$ = $${\frac{{M_x}}{{A}}}$$ = $${\frac{{1}}{{4}}}$$$${\frac{{3}}{{2}}}$$ = $${\frac{{3}}{{8}}}$$

Therefore the centroid is at $$\left(\vphantom{\frac{3}{5},\frac{3}{8}}\right.$$$${\frac{{3}}{{5}}}$$,$${\frac{{3}}{{8}}}$$$$\left.\vphantom{\frac{3}{5},\frac{3}{8}}\right)$$.

Example 8..7   Find the centroid of the region bounded by the curve y = x² and the line y = x.

Remark 8..8   Non-uniform density. Into the calculation of the three integrals you have to incorporate a density function $\sigma$(x, y). The principle remains the same, but to handle the integrals you now need the calculus of severable variables.