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# 1.9: Hemispheres

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

# Uniform solid hemisphere

Figure I.4 will serve. The argument is exactly the same as for the cone. The volume of the elemental slice is $$\pi y^{2} \delta x = \pi (a^{2} - x^{2} ) \delta x$$ and the volume of the hemisphere is $$\frac{2 \pi a^{3}}{3}$$ , so the mass of the slice is

$$M \times \pi (a^{2}-x^{2}) \delta x \div (2 \pi a / 3) = \frac{3M(a^{2}-x^{2}) \delta x }{2a^{3}}$$

where $$M$$ is the mass of the hemisphere. The first moment of mass of the elemental slice is $$x$$ times this, so the position of the centre of mass is

$$\overline{x} = \frac{3}{2a^3} \int_0^a x(a^{2}-x^{2})dx = \frac{3a}{8}$$

# Hollow hemispherical shell.

We may note to begin with that we would expect the centre of mass to be further from the base than for a uniform solid hemisphere.

Again, Figure I.4 will serve. The area of the elemental annulus is $$2 \pi a \delta x$$ (NOT $$2 \pi y \delta x$$!) and the area of the hemisphere is $$2 \pi a^{2}$$ . Therefore the mass of the elemental annulus is

$$M \times 2 \pi a \delta x \div (2 \pi a^{2}) = M \delta x / a$$

The first moment of mass of the annulus is x times this, so the position of the centre of mass is

$$\overline{x} = \int_0^a \frac{xdx}{a} = \frac{a}{2}$$