Modelling with Volumes of Revolution

a) Suggest a suitable value for \(k\).
b) Use your value of \(k\) to estimate the capacity of the tent.
c) State one limitation of this model.

a)
- \(k=10\)

b)
- x intercepts are 100 and -100
- Use volumes of revolution equation
- \(100\pi \int ^{10}_{0} 10^2-y^2\, dx=\frac{200000}{3}\pi\)
- \(\frac{\frac{200000}{3}\pi}{10^3}\)

• What height of sand in the sand timer gives \(40cm^3\)
• \(40cm^3=\pi \int ^k_{0} x^2\, dy\)
• \(40=\pi \int ^k_{0} y^{\frac{2}{3}}\, dx\)
• \(40=\pi\left[ \frac{3}{5}y^{\frac{5}{3}} \right]^k_{0}\)
• \(40=\frac{3}{5}\pi k^{\frac{5}{3}}\)
• \(40=\frac{3}{5}\pi k^{\frac{5}{3}}\)
• \(40\times \frac{5}{3\pi}=21.22\dots\)
• \(21.22\dots=k^{\frac{5}{3}}\)
• \(21.22\dots^3=9556.01\dots\)
• \(\sqrt[5]{ 9556.01\dots }=6.2525\dots\)
• \(k=6.2525\dots\)

• \(18=0.02x^3\)
• \(x=9.65cm\)
• \(2x=19.3\)
  
• \(\pi \int _{0}^{18} \sqrt[3]{ 50y }\, dy=3162.8cm^3\)
  
• Area of paddle = \(\int ^{12}_{0} \sqrt[3]{ 50x }\, dx=75.9\)
• Volume of mixture stired by paddle = \(\pi \int ^{12}_{0} \sqrt[3]{ 50x }^2\, dx=1609.1\)
  
• proportion = \(\frac{1609.1}{3162.8}=0.509\)