Centre of mass by integrationAQA A-Level Further Maths: Revision notes

To find the centre of mass of a uniform lamina under a curve $y=f(x)$ from $x=a$ to $x=b$, split it into thin vertical strips of width $\delta x$. A strip has area $y\,\delta x$, so its mass is proportional to $y\,\delta x$, and its centre of mass is at $(x,\frac12y)$. Taking moments about the axes and letting $\delta x\to0$ replaces the sums by integrals. Because the density is uniform, it cancels, so only areas (or volumes) matter.

For the region between $y=f(x)$, the $x$-axis and $x=a$, $x=b$: $\bar{x}=\frac{\int_a^bxy\,dx}{\int_a^by\,dx},\qquad \bar{y}=\frac{\int_a^b\frac12y^2\,dx}{\int_a^by\,dx}.$ The denominator is the area. Example: $y=x^2$, $0\le x\le2$. Area $=\frac83$, $\int xy\,dx=\int_0^2x^3\,dx=4$ and $\int\frac12y^2\,dx=\frac12\int_0^2x^4\,dx=\frac{16}{5}$, so $\bar{x}=\frac32$ and $\bar{y}=\frac65$. In general for $y=x^2$ from $0$ to $a$: $\bar{x}=\frac{3a}{4}$ and $\bar{y}=\frac{3a^2}{10}$.

Rotate the region through $2\pi$ about the $x$-axis. A thin disc at $x$ has radius $y$, volume $\pi y^2\,\delta x$ and centre of mass on the axis at $x$. Hence $\bar{x}=\frac{\int_a^bxy^2\,dx}{\int_a^by^2\,dx},\qquad \bar{y}=0.$ The $\pi$ cancels, and the denominator is $\frac{V}{\pi}$. By symmetry the centre of mass lies on the axis of rotation. Example: $y=\sqrt{x}$, $0\le x\le4$. $\int_0^4xy^2\,dx=\int_0^4x^2\,dx=\frac{64}{3}$ and $\int_0^4y^2\,dx=\int_0^4x\,dx=8$, so $\bar{x}=\frac83$.

Hemisphere of radius $r$: $y^2=r^2-x^2$ from $0$ to $r$. Numerator $\int_0^rx(r^2-x^2)\,dx=\frac{r^4}{4}$ and denominator $\int_0^r(r^2-x^2)\,dx=\frac{2r^3}{3}$, so $\bar{x}=\frac{3r}{8}$ from the plane face. Cone of height $h$ and base radius $R$: $y=\frac{R}{h}x$, so $\bar{x}=\frac{\int_0^hx^3\,dx}{\int_0^hx^2\,dx}=\frac{3h}{4}$ from the vertex, or $\frac{h}{4}$ from the base. These results can be used within composite bodies, as for a toy made of a hemisphere and a cylinder.

1. Sketch the region and state the limits. 2. Write which formula you are using. 3. Integrate the area (or volume) and each moment separately. 4. Divide and simplify, keeping exact values until the end. 5. Check the answer: $\bar{x}$ must lie between the limits, and $\bar{y}$ must be below the greatest height of the curve. For a region between a curve and the $y$-axis, or a body with an axis of symmetry, use symmetry to find one coordinate without integration.