Centres of mass of uniform bodies by integrationEdexcel International A Level Further Maths: Revision notes

The centre of mass of a body is the point where its whole weight can be taken to act. For a system of parts with masses $m_i$ at $x_i$, $\left(\sum m_i\right)\bar x=\sum m_ix_i$, and similarly for $\bar y$. For a uniform body the mass is proportional to length (rod), area (lamina) or volume (solid), so use those in place of mass. A hole or removed piece counts as a negative mass. Example: a circular lamina of radius $4$ with centre at the origin has a circular hole of radius $1$ with centre $(2,0)$. Areas are $16\pi$ and $\pi$, so $15\pi\,\bar x=16\pi(0)-\pi(2)$ and $\bar x=-\frac{2}{15}$, $\bar y=0$.

If a uniform body has an axis or plane of symmetry, the centre of mass lies on it. The formulae booklet gives results you can quote, including:

• triangular lamina: $\frac23$ of the way along a median from a vertex
• semicircular lamina: $\frac{4r}{3\pi}$ from the diameter
• solid hemisphere: $\frac{3r}{8}$ from the plane face; hemispherical shell: $\frac r2$
• solid cone or pyramid: $\frac14$ of the height from the base; conical shell: $\frac13$ The questions that ask you to use integration want these derived from scratch, so quote a booklet result only if the question does not say 'show' or 'use integration'. To combine a hemisphere and a cylinder, use volumes $\frac23\pi r^3$ and $\pi r^2h$ as masses.

For a uniform lamina under $y=f(x)$ from $x=a$ to $x=b$, with area $A=\int_a^by\,dx$: $\bar x=\frac{\int_a^bxy\,dx}{A},\qquad\bar y=\frac{\frac12\int_a^by^2\,dx}{A}.$ The strip has width $dx$, its centre of mass is at $x$ horizontally and at height $\frac y2$. Example: $y=\sqrt x$, $0\le x\le4$. $A=\int_0^4\sqrt x\,dx=\frac{16}{3}$. $\int_0^4x^{3/2}\,dx=\frac{64}{5}$ so $\bar x=\frac{64/5}{16/3}=\frac{12}{5}$. $\frac12\int_0^4x\,dx=4$ so $\bar y=\frac{4}{16/3}=\frac34$.

For a uniform solid formed by rotating $y=f(x)$ about the $x$-axis from $x=a$ to $x=b$, each disc has volume $\pi y^2\,dx$ and its centre of mass is on the axis at $x$: $\bar x=\frac{\int_a^bxy^2\,dx}{\int_a^by^2\,dx},$ because the factor $\pi$ cancels. The centre of mass is on the $x$-axis. Example: a cone from $y=\frac rhx$, $0\le x\le h$. $\bar x=\frac{\int_0^hx^3\,dx}{\int_0^hx^2\,dx}=\frac{h^4/4}{h^3/3}=\frac{3h}{4}$ from the vertex, so $\frac h4$ from the base. For rotation about the $y$-axis: $\bar y=\frac{\int yx^2\,dy}{\int x^2\,dy}$. A hemisphere: $x^2=r^2-y^2$, $0\le y\le r$, so $\bar y=\frac{r^4/4}{2r^3/3}=\frac{3r}{8}$.

Often one part of a composite body needs integration and the rest is standard. Find each part's mass (area or volume) and centre of mass, then take moments about a convenient axis through the end or the origin. If a cylinder of radius $r$ and height $h$ is attached to the base of a body, the cylinder's centre of mass is $\frac h2$ from the joint, with mass proportional to $\pi r^2h$. Express the result in exact form first, then give the decimal required. If a body is asked to be 'removed', subtract. Always state from which point or axis your distance is measured.