Minimizing Resultant Volume

Calculus Level pending

Let a a , b b , c c , and, d d be real constants. Minimize the volume of the region bounded between y = x 4 + a x 3 + b x 2 + c x + d 1 − x 2 4 y = \dfrac {x^4 + ax^3 +bx^2 + cx + d} {\sqrt[4]{1 - x^2}} , x = − 1 x= -1 and x = 1 x= 1 , when it is revolved about the x x -axis.

If this volume can be expressed as m n π 2 \dfrac mn \pi^2 , where m m and n n are coprime positive integers, submit your answer as m + n m+n .


The answer is 129.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Nov 14, 2016

The Volume V ( a , b , c , d ) = π ∗ ∫ − 1 1 ( x 4 + a x 3 + b x 2 + c x + d ) 2 1 − x 2 d x = {\bf V(a,b,c,d) = \pi * \int_{-1}^{1} \dfrac {(x^4 + ax^3 +bx^2 + cx + d)^2} {\sqrt{1 - x^2}} } \: dx=

π ∗ ∫ − 1 1 ( x 8 + 2 a ∗ x 7 + ( a 2 + 2 b ) ∗ x 6 + 2 ( a b + c ) ∗ x 5 + ( 2 d + 2 a c + b 2 ) ∗ x 4 + 2 ( a d + b c ) ∗ x 3 + ( 2 b d + c 2 ) ∗ x 2 + 2 c d ∗ x + d 2 ) 1 − x 2 d x = {\bf \pi * \int_{-1}^{1} \dfrac {(x^8 + 2a * x^7 + (a^2 + 2b) * x^6 + 2(ab + c) * x^5 + (2d + 2ac + b^2) * x^4 + 2(ad + bc) * x^3 + (2bd + c^2) * x^2 + 2cd * x + d^2) } {\sqrt{1 - x^2}} } \: dx =

Let x = s i n θ ⟹ d x = c o s θ d θ ⟹ {\bf x = sin\theta \implies dx = cos\theta \: d\theta \implies }

V ( a , b , c , d ) = π ∗ ∫ − π 2 π 2 ( ( 1 − c o s ( 2 θ ) 2 ) 4 + 2 a ( 1 − c o s 2 θ ) 3 s i n θ + ( a 2 + 2 b ) ( 1 − c o s ( 2 θ ) 2 ) 3 + {\bf V(a,b,c,d) = \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} ((\frac{1 - cos(2\theta)}{2})^4 + 2a(1 - cos^2\theta)^3 sin\theta + (a^2 + 2b) (\frac{1 - cos(2\theta)}{2})^3 + } 2 ( a b + c ) ( 1 − c o s 2 θ ) 2 s i n θ + ( 2 d + 2 a c + b 2 ) ( 1 − c o s ( 2 θ ) 2 ) 2 + 2 ( a d + b c ) ( 1 − c o s 2 θ ) s i n θ + {\bf 2(ab + c) (1 - cos^2\theta)^2 sin\theta + (2d + 2ac + b^2) (\frac{1 - cos(2\theta)}{2})^2 + 2(ad + bc) (1 - cos^2\theta) sin\theta + } ( 2 b d + c 2 ) ( 1 − c o s ( 2 θ ) 2 ) + {\bf (2bd + c^2) (\frac{1 - cos(2\theta)}{2}) + } 2 c d s i n θ + d 2 ) d θ . {\bf 2cd sin\theta + d^2) \: d\theta. }

After simplifying we obtain:

I 1 = π 16 ∗ ∫ − π 2 π 2 ( 1 − c o s ( 2 θ ) 2 ) 4 d θ = 35 128 π 2 {\bf I_{1} = \frac{\pi}{16} * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (\frac{1 - cos(2\theta)}{2})^4 \: d\theta = \frac{35}{128} \pi^2 }

I 2 = 2 a π ∗ ∫ − π 2 π 2 ( 1 − c o s 2 θ ) 3 s i n θ d θ = 0 {\bf I_{2} = 2a \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (1 - cos^2\theta)^3 sin\theta \: d\theta = 0}

I 3 = ( a 2 + 2 b 8 ) π ∗ ∫ − π 2 π 2 ( 1 − c o s ( 2 θ ) ) 3 d θ = 5 16 ( a 2 + 2 b ) π 2 {\bf I_{3} = (\frac{a^2 + 2b}{8}) \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (1 - cos(2\theta))^3 \: d\theta = \frac{5}{16} (a^2 + 2b) \pi^2 }

I 4 = 2 ( a b + c ) π ∗ ∫ − π 2 π 2 ( 1 − c o s 2 θ ) 2 s i n θ d θ = 0 {\bf I_{4} = 2(ab + c) \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (1 - cos^2\theta)^2 sin\theta \: d\theta = 0 }

I 5 = ( 2 d + 2 a c + b 2 4 ) π ∗ ∫ − π 2 π 2 ( 1 − c o s ( 2 θ ) ) 2 d θ = 3 8 ( b 2 + 2 a c + 2 d ) π 2 {\bf I_{5} = (\frac{2d + 2ac + b^2}{4}) \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (1 - cos(2\theta))^2 \: d\theta = \frac{3}{8} (b^2 + 2ac + 2d) \pi^2 }

I 6 = 2 ( a d + b c ) π ∗ ∫ − π 2 π 2 ( 1 − c o s 2 θ ) s i n θ d θ = 0 {\bf I_{6} = 2(ad + bc) \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (1 - cos^2\theta) sin\theta \: d\theta = 0 } , I 6 = I 4 = I 2 = 0 {\bf I_{6} = I_{4} = I_{2} = 0 }

I 7 = 2 b d + c 2 2 π ∗ ∫ − π 2 π 2 ( 1 − c o s ( 2 θ ) ) d θ = ( 2 b d + c 2 2 ) π 2 {\bf I_{7} = \frac{2bd + c^2}{2} \pi * \int_{\frac {-\pi}{2} }^{\frac{\pi}{2}} (1 - cos(2\theta)) \: d\theta = (\frac{2bd + c^2}{2}) \pi^2 }

I 8 = d 2 π 2 {\bf I_{8} = d^2 \pi^2 }

⟹ {\bf \implies }

V ( a , b , c , d ) = π 2 ∗ ( 35 128 + 5 16 ( a 2 + 2 b ) + 3 8 ( b 2 + 2 a c + 2 d ) + 1 2 ( 2 b d + c 2 ) + d 2 ) {\bf V(a,b,c,d) = \pi^2 * (\frac{35}{128} +\frac{5}{16}(a^2 + 2b) + \frac{3}{8} (b^2 + 2ac + 2d) + \frac{1}{2} (2bd + c^2) + d^2) }

⟹ {\bf \implies }

∂ V ∂ a = 0 ⟹ 5 a + 6 c = 0 {\bf \frac{\partial {V}}{\partial {a}} = 0 \implies 5a + 6c = 0 }

∂ V ∂ c = 0 ⟹ 3 a + 4 c = 0 {\bf \frac{\partial {V}}{\partial {c}} = 0 \implies 3a + 4c = 0 }

⟹ a = c = 0 {\bf \implies a = c = 0 }

∂ V ∂ b = 0 ⟹ 6 b + 8 d = − 5 {\bf \frac{\partial {V}}{\partial {b}} = 0 \implies 6b + 8d = -5 }

∂ V ∂ d = 0 ⟹ 4 b + 8 d = − 3 {\bf \frac{\partial {V}}{\partial {d}} = 0 \implies 4b + 8d = -3 }

⟹ b = − 1 , d = 1 8 {\bf \implies b = -1, d = \frac{1}{8} }

and, ∂ 2 V ∂ a 2 = 5 8 π 2 > 0 {\bf \frac{\partial^2V}{\partial a^2} = \frac{5}{8} \pi^2 > 0 }

∂ 2 V ∂ b 2 = 3 4 π 2 > 0 {\bf \frac{\partial^2V}{\partial b^2} = \frac{3}{4} \pi^2 > 0 }

∂ 2 V ∂ c 2 = π 2 > 0 {\bf \frac{\partial^2V}{\partial c^2} = \pi^2 > 0 }

∂ 2 V ∂ d 2 = 2 π 2 > 0 {\bf \frac{\partial^2V}{\partial d^2} = 2 \pi^2 > 0 }

and, the Hessian matrix

M = π 2 ∗ [ 5 8 0 0 0 0 3 4 0 0 0 0 1 0 0 0 0 2 ] {\bf M = \pi^2 * [ \begin{array}{cccc} \frac{5}{8} & 0 & 0 & 0 \\ 0 & \frac{3}{4} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} ] }

d e t ( M ) > 0 ⟹ {\bf det(M) > 0 \implies }

we have a minimum value at ( 0 , − 1 , 0 , 1 8 ) {\bf (0, -1, 0, \frac{1}{8}) }

⟹ V = 1 128 π 2 = m n π 2 ⟹ {\bf \implies V = \dfrac{1}{128} \pi^2 = \dfrac mn \pi^2 \implies }

m + n = 129. {\bf m + n = 129. }

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Haha, this is the exact same question as @Guillermo Templado 's problem !

Pi Han Goh - 4 years, 7 months ago

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I'm going to continue with my problem, the next two propositions are impressive... Hold on , they're coming...

Guillermo Templado - 4 years, 7 months ago

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