\bar{y} for a lamina under y=f(x)?

\bar{y}=\frac{\int\frac12y^2\,dx}{\int y\,dx}

\bar{x} for a solid of revolution about the x-axis?

\bar{x}=\frac{\int xy^2\,dx}{\int y^2\,dx}

Why does \pi cancel for a solid of revolution?

It appears in both the numerator and the denominator.

\bar{y} for a solid of revolution about the x-axis?

0, because the centre of mass lies on the axis of rotation.

Centre of mass of a hemisphere of radius r?

\frac{3r}{8} from the centre of the plane face.

Centre of mass of a cone of height h?

\frac{3h}{4} from the vertex, \frac{h}{4} from the base.

For y=x^2 from 0 to a, find \bar{x} and \bar{y}.

\bar{x}=\frac{3a}{4}, \bar{y}=\frac{3a^2}{10}

Check on a centre of mass found by integration?

\bar{x} lies between the limits and \bar{y} is below the greatest height.

Centre of mass by integrationAQA A-Level Further Maths: Flashcards

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$\bar{x}$ for a lamina under $y=f(x)$?

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$\bar{x}$ for a lamina under $y=f(x)$ ?: $\bar{x}=\frac{\int xy\,dx}{\int y\,dx}$
$\bar{y}$ for a lamina under $y=f(x)$ ?: $\bar{y}=\frac{\int\frac12y^2\,dx}{\int y\,dx}$
Where is the centre of mass of a thin strip of height $y$ ?: At $(x,\frac12y)$ .
What does the denominator represent for a lamina?: The area of the region.
$\bar{x}$ for a solid of revolution about the $x$ -axis?: $\bar{x}=\frac{\int xy^2\,dx}{\int y^2\,dx}$
Why does $\pi$ cancel for a solid of revolution?: It appears in both the numerator and the denominator.
$\bar{y}$ for a solid of revolution about the $x$ -axis?: $0$ , because the centre of mass lies on the axis of rotation.
Volume of a solid of revolution about the $x$ -axis?: $V=\pi\int y^2\,dx$
Centre of mass of a hemisphere of radius $r$ ?: $\frac{3r}{8}$ from the centre of the plane face.
Centre of mass of a cone of height $h$ ?: $\frac{3h}{4}$ from the vertex, $\frac{h}{4}$ from the base.
For $y=x^2$ from $0$ to $a$ , find $\bar{x}$ and $\bar{y}$ .: $\bar{x}=\frac{3a}{4}$ , $\bar{y}=\frac{3a^2}{10}$
Check on a centre of mass found by integration?: $\bar{x}$ lies between the limits and $\bar{y}$ is below the greatest height.

Exam questions on Centre of mass by integration

A uniform lamina occupies the region $R$ bounded by the curve $y=x^2$ , the $x$ -axis and the line $x=2$ . Lengths are in metres.
A second uniform lamina occupies the region bounded by $y=x^2$ , the $x$ -axis and the line $x=3$ . Find its $x$ -coordinate of the centre of mass.2 marks
The region $R$ is bounded by the curve $y=\sqrt{x}$ , the $x$ -axis and the line $x=4$ . A uniform solid is formed by rotating $R$ through $2\pi$ radians about the $x$ -axis. Lengths are in centimetres.
A second solid is formed by rotating the region bounded by $y=\sqrt{x}$ , the $x$ -axis and the line $x=9$ about the $x$ -axis. Find the $x$ -coordinate of its centre of mass.2 marks
A uniform lamina occupies the region bounded by the curve $y=4-x^2$ , the $x$ -axis and the $y$ -axis, for $0\le x\le2$ . Lengths are in metres.
Show that the $x$ -coordinate of the centre of mass of the lamina is $\frac34$ m.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).