A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.
(a) Illustrate this information in a diagram.
(b) If the volume of the reservoir is (333frac{1}{3}pi m^{3}), calculate, correct to the nearest whole number, the :
(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take (pi = frac{22}{7})].
Explanation
(a)
(i) Using volume of reservoir = volume of hemisphere + volume of cone.
(implies 333frac{1}{3}pi = frac{2}{3}pi r^{3} + frac{1}{3}pi r^{2} h)
(implies frac{1000}{3}pi = frac{1}{3}pi (2r^{3} + r^{2}h))
Let the radius of the hemisphere be x m;
Then (frac{1000}{3} = frac{1}{3}(2x^{3} + x^{2}(6x)))
(1000 = 2x^{3} + 6x^{3})
(1000 = 8x^{3} implies x^{3} = frac{1000}{8} = 125)
(x = sqrt[3]{125} = 5 m)
Hence, the volume of the hemisphere = (frac{2}{3}pi r^{3})
(where r = x = 5 m)
= (frac{2}{3} times frac{22}{7} times 5^{3} m^{3})
= (frac{5500}{21} m^{3} approxeq 261.905 m^{3})
= (262 m^{3} ) (nearest whole number)
(ii) Total surface area of reservoir
= surface area of hemisphere + surface area of cone
= (2pi r^{2} + pi rl)
From the diagram above, (l^{2} = 30^{2} + 5^{2})
(l^{2} = 900 + 25 = 925)
(l = sqrt{925} = 30.41 m)
Hence, the total surface area of the reservoir = (2 times frac{22}{7} times 5 times 5 + frac{22}{7} times 5 times 30.41)
= (157.143 + 477.871)
= (635.104 m^{2})
(approxeq 635 m^{2}) (to the nearest whole number).