Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 12

Solution

d V = r d r d θ d z

0 \leq θ < 2 π

0 \leq z \leq h

and

0 \leq r \leq a (1 - z / h)

s = 1 - z / h

. Integrating first over radius gives the mass per unit height

\frac{d M}{d z} = 2 π ρ_{0} [\frac{a ^{2} s ^{2}}{2} + \frac{a ^{2} s ^{3}}{3}] = π ρ_{0} a^{2} (s^{2} + \frac{2}{3} s^{3}) .

Therefore

M = π ρ_{0} a^{2} h \int_{0}^{1} (s^{2} + \frac{2}{3} s^{3}) d s = \frac{1}{2} π ρ_{0} a^{2} h .

For the first moment about the base, use

z = h (1 - s)

M d = π ρ_{0} a^{2} h^{2} \int_{0}^{1} (1 - s) (s^{2} + \frac{2}{3} s^{3}) d s = \frac{7}{60} π ρ_{0} a^{2} h^{2} .

Dividing by the total mass gives

d = \frac{7 h}{30} .

h /4

, while the component proportional to

r

h /5

and half as much mass. Their mass-weighted mean is

7 h /30

. Density increasing toward the wider outer parts consequently moves the center below the uniform cone's

h /4

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