Volume of solids of revolution

Calculus Level pending

Let a {\bf a } and b {\bf b } be real positive constants.

The volume of the region bounded by the curve x = a b b 2 y 2 , y = 0 a n d y = b {\bf x = \frac{a}{b} * \sqrt{b^2 - y^2}, y = 0 \: and \: y = b } , when revolved about the line x = a {\bf x = a } , can be expressed as a 2 b π ( m n π p ) , {\bf a^2b\pi * (\frac{m}{n} - \frac{\pi}{p}) , } where m , {\bf m, } n {\bf n } and p {\bf p } are coprime positive integers.

Find m + n + p . {\bf m + n + p }.


The answer is 10.

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1 solution

Rocco Dalto
Nov 11, 2016

The volume V = π 0 b ( a x ) 2 d y = {\bf V = \pi * \int_{0}^{b} (a - x)^2 dy = }

a 2 b 2 π 0 b 2 b 2 y 2 2 b b 2 y 2 d y = {\bf \frac{a^2}{b^2}\pi * \int_{0}^{b} 2b^2 - y^2 -2b\sqrt{b^2 - y^2} dy = }

a 2 b 2 π ( 2 b 2 y y 3 3 ) 0 b 2 b 0 b b 2 y 2 d y = {\bf \frac{a^2}{b^2}\pi * (2b^2 y - \frac{y^3}{3})|_{0}^{b} - 2b \int_{0}^{b} \sqrt{b^2 - y^2} dy = }

5 3 a 2 b π 2 a 2 π b 0 b b 2 y 2 d y {\bf \frac{5}{3} a^2 b \pi * - \frac{2 a^2\pi}{b} * \int_{0}^{b} \sqrt{b^2 - y^2} dy}

For I = 0 b b 2 y 2 {\bf I = \int_{0}^{b} \sqrt{b^2 - y^2} }

Let y = b s i n θ d y = b c o s θ d θ {\bf y = b sin\theta \implies dy = b cos\theta d\theta \implies }

I = b 2 2 0 π 2 ( 1 + c o s ( 2 θ ) ) d θ = {\bf I =\frac{b^2}{2} \int_{0}^{\frac{\pi}{2}} (1 + cos(2\theta)) d\theta = }

b 2 2 ( θ + 1 2 s i n ( 2 θ ) ) 0 π 2 = {\bf \frac{b^2}{2} (\theta + \frac{1}{2} sin(2\theta))|_{0}^{\frac{\pi}{2}} = } π 4 b 2 {\bf \frac{\pi}{4} b^2 \implies }

V ( a , b ) = 5 3 a 2 b π a 2 b π 2 2 = {\bf V(a,b) = \frac{5}{3} a^2 b \pi - \frac{a^2 b \pi^2}{2} = } a 2 b π ( 5 3 π 2 ) {\bf a^2 b \pi * (\frac{5}{3} - \frac{\pi}{2}) } = a 2 b π ( m n π p ) {\bf = a^2b\pi * (\frac{m}{n} - \frac{\pi}{p}) }

m + n + p = 10. {\bf \implies m + n + p = 10. }

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