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 = 3 b {\bf x = \frac{a}{b} * \sqrt{b^2 + y^2}, y = 0 \: and \: y = \sqrt{3} * b } , when revolved about the line x = a {\bf x = a } , can be expressed as a 2 b π ( m l n ( n + m ) ) , {\bf a^2b\pi * (\sqrt{m} - ln(n + \sqrt{m})), } where m {\bf m } and n {\bf n } are coprime positive integers.

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


The answer is 5.

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1 solution

Rocco Dalto
Nov 11, 2016

The volume V = π 0 3 b ( x a ) 2 d y = {\bf V = \pi * \int_{0}^{\sqrt{3} * b} (x - a)^2 dy = }

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

Let I 2 = 2 b 0 3 b b 2 + y 2 d y {\bf I_{2} = 2b * \int_{0}^{\sqrt{3} * b} \sqrt{b^2 + y^2} dy' }

Using trig substitution, Let y = b t a n θ d y = b s e c 2 θ {\bf y = b tan\theta \implies dy = b sec^2\theta \implies}

I 2 = 2 b 3 0 π 3 s e c 3 ( θ ) d θ {\bf I_{2} = 2b^3 * \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta }

Using integration by parts let u = s e c θ d u = s e c θ t a n θ {\bf u = sec\theta \implies du = sec\theta tan\theta}

d v = s e c 2 ( θ ) v = t a n θ {\bf dv = sec^2(\theta) \implies v = tan\theta \implies }

0 π 3 s e c 3 ( θ ) d θ = s e c θ t a n θ 0 π 3 0 π 3 s e c 3 ( θ ) d θ + 0 π 3 s e c θ d θ {\bf \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta = sec\theta tan\theta|_{0}^{\frac{\pi}{3}} - \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta + \int_{0}^{\frac{\pi}{3}} sec\theta d\theta \implies }

2 0 π 3 s e c 3 ( θ ) d θ = ( s e c θ t a n θ + l n s e c θ + t a n θ ) 0 π 3 {\bf 2 * \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta = (sec\theta tan\theta + ln|sec\theta + tan\theta|)|_{0}^{\frac{\pi}{3}} \implies }

0 π 3 s e c 3 ( θ ) d θ = 1 2 ( s e c θ t a n θ + l n s e c θ + t a n θ ) 0 π 3 = {\bf \int_{0}^{\frac{\pi}{3}} sec^3(\theta) d\theta = \frac{1}{2} * (sec\theta tan\theta + ln|sec\theta + tan\theta|)|_{0}^{\frac{\pi}{3}} = }

1 2 ( 2 3 + l n ( 2 + 3 ) ) {\bf \frac{1}{2} * (2\sqrt{3} + ln(2 + \sqrt{3})) \implies }

I 2 = b 3 ( 2 3 + l n ( 2 + 3 ) ) {\bf I_{2} = b^3 * (2\sqrt{3} + ln(2 + \sqrt{3})) }

Let I 1 = 0 3 b 2 b 2 + y 2 d y {\bf I_{1} = \int_{0}^{\sqrt{3} * b} 2b^2 + y^2 dy \implies }

I 1 = 2 b 2 y + y 3 3 0 3 b = 3 3 b 3 {\bf I_{1} = 2b^2 y + \frac{y^3}{3}|_{0}^{\sqrt{3} * b} = 3\sqrt{3} b^3 \implies}

V = a 2 b π ( 3 l n ( 2 + 3 ) ) = a 2 b π ( m l n ( n + m ) ) {\bf V = a^2 b \pi * (\sqrt{3} - ln(2 + \sqrt{3})) = a^2b\pi * (\sqrt{m} - ln(n + \sqrt{m})) \implies }

m + n = 5. {\bf m + n = 5. }

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