Volume Of Solids of Revolution

Calculus Level 4

Let β \beta be a positive integer and ∣ x ∣ > 1 |x| > 1 .

Let f ( x ) = ∑ j = 0 ∞ x β + j + 1 − x β + j x β + j ∗ x β + j + 1 f(x) = \sum_{j = 0}^{\infty} \dfrac{x^{\beta + j + 1} - x^{\beta + j}}{x^{\beta + j} * x^{\beta + j + 1}} and g ( x ) = 1 a x + b g(x) = \dfrac{1}{ax + b} .

Let V f V_{f} be the volume of the region formed when the curve f ( x ) f(x) is revolved about the x x -axis on [ 2 , ∞ ) [2,\infty) and V g V_{g} be the volume of the region formed when the curve g ( x ) g(x) is revolved about the x x -axis on [ 2 , ∞ ) [2,\infty)

If f ( 2 ) = g ( 2 ) f(2) = g(2) and V f = V g V_{f} = V_{g} and the volume V V of the region bounded by the curves f ( x ) f(x) and g ( x ) g(x) when revolved about the x x -axis on ( − ∞ , − 2 ] (-\infty,-2] can be expressed as V = α α ( β − w ) π ( α β − w ) ( α α β − λ ) ( α α β − w ) V = \dfrac{\alpha^{\alpha} (\beta - w)\pi}{(\alpha\beta - w)(\alpha^{\alpha}\beta - \lambda) (\alpha^{\alpha\beta - w})} , where α , w \alpha, w and λ \lambda are coprime positive integers, find α + w + λ \alpha + w + \lambda .

I used β = 2 \beta = 2 for the above graph.


The answer is 6.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jul 22, 2018

Let ∣ x ∣ > 1 |x| > 1 .

f ( x ) = ∑ j = 0 ∞ x β + j + 1 − x β + j x β + j ∗ x β + j + 1 = ∑ j = 0 ∞ x β + j ( x − 1 ) x 2 ( β + j ) ∗ x = x − 1 x β + 1 ∑ j = 0 ∞ ( 1 x ) j = ( x − 1 x β + 1 ) ( x x − 1 ) = 1 x β f(x) = \sum_{j = 0}^{\infty} \dfrac{x^{\beta + j + 1} - x^{\beta + j}}{x^{\beta + j} * x^{\beta + j + 1}} = \sum_{j = 0}^{\infty} \dfrac{x^{\beta + j} (x - 1)}{x^{2(\beta + j)} * x} = \dfrac{x - 1}{x^{\beta + 1}} \sum_{j = 0}^{\infty} (\dfrac{1}{x})^{j} = (\dfrac{x - 1}{x^{\beta + 1}}) (\dfrac{x}{x - 1}) = \boxed{\dfrac{1}{x^{\beta}}}

Let g ( x ) = 1 a x + b g(x) = \dfrac{1}{ax + b} .

f ( 2 ) = g ( 2 ) ⟹ 2 a + b = 2 β f(2) = g(2) \implies \boxed{2a + b = 2^{\beta}}

V f = π ∫ 2 ∞ ( f ( x ) ) 2 d x = π ∫ 2 ∞ x − 2 β d x = − π 2 β − 1 ( 1 x 2 β − 1 ) ∣ 2 ∞ = V_{f} = \pi\int_{2}^{\infty} (f(x))^2 dx = \pi\int_{2}^{\infty} x^{-2\beta} dx = -\dfrac{\pi}{2\beta - 1}(\dfrac{1}{x^{2\beta - 1}})|_{2}^{\infty} = π ( 2 β − 1 ) 2 2 β − 1 \dfrac{\pi}{(2\beta - 1)2^{2\beta - 1}}

⟹ π ( 2 β − 1 ) 2 2 β − 1 = V g = π ∫ ( g ( x ) ) 2 d x = π a ∫ 2 ∞ ( a x + b ) − 2 ∗ a d x = − π a ( 1 a x + b ) ∣ 2 ∞ = π a ( 2 a + b ) \implies \dfrac{\pi}{(2\beta - 1)2^{2\beta - 1}} = V_{g} = \pi\int (g(x))^2 dx = \dfrac{\pi}{a}\int_{2}^{\infty} (ax + b)^{-2} * a \:\ dx = -\dfrac{\pi}{a}(\dfrac{1}{ax + b})|_{2}^{\infty} = \dfrac{\pi}{a(2a + b)}

⟹ a ( 2 a + b ) = ( 2 β − 1 ) 2 2 β − 1 \implies \boxed{a(2a + b) = (2\beta - 1) 2^{2\beta - 1}}

and

2 a + b = 2 β \boxed{2a + b = 2^{\beta}}

Solving the system we obtain:

a = ( 2 β − 1 ) 2 β − 1 a = (2\beta - 1)2^{\beta - 1} and b = 2 β + 1 ( 1 − β ) ⟹ g ( x ) = 1 2 β − 1 ( ( 2 β − 1 ) x − 4 ( β − 1 ) ) b = 2^{\beta + 1} (1 - \beta) \implies g(x) = \dfrac{1}{2^{\beta - 1}((2\beta - 1)x - 4(\beta - 1))}

V 1 ( β ) = π ∫ − ∞ − 2 ( f ( x ) ) 2 d x = π ( 2 β − 1 ) 2 2 β − 1 V_{1}(\beta) = \pi\int_{-\infty}^{-2} (f(x))^2 dx = \dfrac{\pi}{(2\beta - 1)2^{2\beta - 1}}

and

V 2 ( β ) = π ∫ − ∞ − 2 ( g ( x ) ) 2 d x = − π a ( 1 a x + b ) ∣ 2 ∞ = π a ( 2 a − b ) = π ( 2 β − 1 ) 2 2 β − 1 ( 4 β − 3 ) V_{2}(\beta) = \pi\int_{-\infty}^{-2} (g(x))^2 dx = -\dfrac{\pi}{a}(\dfrac{1}{ax + b})|_{2}^{\infty} = \dfrac{\pi}{a(2a - b)} = \dfrac{\pi}{(2\beta - 1)2^{2\beta - 1}(4\beta - 3)}

⟹ V 1 ( β ) − V 2 ( β ) = 4 ( β − 1 ) π ( 2 β − 1 ) ( 4 β − 3 ) 2 2 β − 1 = \implies V_{1}(\beta) - V_{2}(\beta) = \dfrac{4(\beta - 1)\pi}{(2\beta - 1)(4\beta - 3) 2^{2\beta - 1}} = 2 2 ( β − 1 ) π ( 2 β − 1 ) ( 2 2 β − 3 ) 2 2 β − 1 = α α ( β − w ) π ( α β − w ) ( α α β − λ ) ( α α β − w ) ⟹ α + w + λ = 6 \dfrac{2^2(\beta - 1)\pi}{(2\beta - 1)(2^2\beta - 3) 2^{2\beta - 1}} = \dfrac{\alpha^{\alpha} (\beta - w)\pi}{(\alpha\beta - w)(\alpha^{\alpha}\beta - \lambda) (\alpha^{\alpha\beta - w})} \implies \alpha + w + \lambda = \boxed{6} .

Note: For β = 2 ⟹ f ( x ) = 1 x 2 \beta = 2 \implies f(x) = \dfrac{1}{x^2} and g ( x ) = 1 6 x − 8 ⟹ V 1 ( 2 ) − V 2 ( 2 ) = π 30 g(x) = \dfrac{1}{6x - 8} \implies V_{1}(2) - V_{2}(2) = \dfrac{\pi}{30} .

For β = 3 ⟹ f ( x ) = 1 x 3 \beta = 3 \implies f(x) = \dfrac{1}{x^3} and g ( x ) = 1 20 x − 32 ⟹ V 1 ( 3 ) − V 2 ( 3 ) = π 180 g(x) = \dfrac{1}{20x - 32} \implies V_{1}(3) - V_{2}(3) = \dfrac{\pi}{180} .

For trivial case β = 1 ⟹ f ( x ) = g ( x ) = 1 / x \beta = 1 \implies f(x) = g(x) = 1/x and V 1 ( 1 ) = V 2 ( 1 ) = π 2 V_{1}(1) = V_{2}(1) = \dfrac{\pi}{2} .

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