More on Volumes of Revolution

Calculus Level 3

If the volume V V of the region bounded by f ( x ) = 4 5 ( x + 1 x 2 ) f(x) = \dfrac{4}{5}(\sqrt{x} + \sqrt{1 - x^2}) and g ( x ) = 8 5 x g(x) = \dfrac{8}{5}\sqrt{x} when revolved about the y y axis is V = a 3 π b ( c d ( b c a ) d a ( a + d b d b ) ) V = \dfrac{a^3\pi}{b}(\dfrac{c}{d} - (\dfrac{\sqrt{b} - c}{a})^{\dfrac{d}{a}} (\dfrac{a + d\sqrt{b}}{db})) , where a , b , c a,b,c and d d are coprime positive integers, find a + b + c + d a + b + c + d .


The answer is 11.

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1 solution

Rocco Dalto
Mar 10, 2018

f ( x ) = g ( x ) x 2 + x 1 = 0 x = 5 1 2 f(x) = g(x) \implies x^2 + x - 1 = 0 \implies x = \dfrac{\sqrt{5} - 1}{2} \implies the two curves intersect at x = 5 1 2 V = 2 π 0 5 1 2 x ( 4 5 ( x + 1 x 2 ) 8 5 x ) d x = x = \dfrac{\sqrt{5} - 1}{2} \implies V = 2\pi \int_{0}^{\frac{\sqrt{5} - 1}{2}} x(\dfrac{4}{5}(\sqrt{x} + \sqrt{1 - x^2}) - \dfrac{8}{5}\sqrt{x}) dx = 8 π 5 0 5 1 2 x 1 x 2 x 3 2 d x = \dfrac{8\pi}{5} \int_{0}^{\frac{\sqrt{5} - 1}{2}} x\sqrt{1 - x^2} - x^{\frac{3}{2}} dx = 8 π 5 ( 1 3 ( 1 x 2 ) 3 2 2 5 x 5 2 ) 0 5 1 2 = \dfrac{8\pi}{5} (\dfrac{-1}{3} (1 - x^2)^{\frac{3}{2}} - \dfrac{2}{5} x^{\frac{5}{2}})|_{0}^{\frac{\sqrt{5} - 1}{2}} =

8 π 5 ( 1 3 ( 5 1 2 ) 3 2 ( 2 + 3 5 15 ) ) = \dfrac{8\pi}{5} (\dfrac{1}{3} - (\dfrac{\sqrt{5} - 1}{2})^{\dfrac{3}{2}} (\dfrac{2 + 3\sqrt{5}}{15})) = a 3 π b ( c d ( b c a ) d a ( a + d b d b ) ) \dfrac{a^3\pi}{b}(\dfrac{c}{d} - (\dfrac{\sqrt{b} - c}{a})^{\dfrac{d}{a}} (\dfrac{a + d\sqrt{b}}{db})) a + b + c + d = 11 \implies a + b + c + d = \boxed{11} .

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