Infinite Volume Or Finite Volume?

Calculus Level 4

Rotate the curve drawn by the function f ( x ) = x − a f(x)=x^{-a} , a a a positive real number, for x > 1 x>1 , about the x x -axis, as so:

For which values of a a is the volume of the resulting solid finite?

Bonus: For the correct inequality, what is the volume in terms of a a ?

a > 1 a>1 a > 0 a>0 a ≥ 1 a\geq1 a > 1 2 a>\frac{1}{2} a ≥ 1 2 a\geq\frac{1}{2}

Andrei Li by Andrei Li , 16, Canada

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1 solution

Andrei Li
Aug 17, 2018

The area can be expressed as the area of an infinite number of discs, with 'height' Δ x \Delta x , 'radius' x − a = 1 x a x^{-a}=\frac{1}{x^a} , and 'area' π x − 2 a Δ x \pi x^{-2a} \Delta x . As Δ x → 0 \Delta x \to 0 , the limit is the improper integral

lim ⁡ x → ∞ π ∫ 1 ∞ x − 2 a d x \lim_{x\to\infty} \pi \int_1^{\infty} x^{-2a} dx

Using the power rule for integration, the integral becomes

∫ x − 2 a d x = x 1 − 2 a 1 − 2 a = 1 ( 1 − 2 a ) x 2 a − 1 \int x^{-2a} dx = \frac{x^{1-2a}}{1-2a}= \frac{1}{(1-2a)x^{2a-1}}

When x = 1 x=1 , this is π 1 − 2 a \frac{\pi}{1-2a} . Now, in order for the volume to be finite, lim ⁡ x → ∞ 1 ( 1 − 2 a ) x 2 a − 1 \lim_{x\to\infty}\frac{1}{(1-2a)x^{2a-1}} must be equal to 0 0 , or

lim ⁡ x → ∞ ( 1 − 2 a ) x 2 a − 1 = ∞ \lim_{x\to\infty}(1-2a)x^{2a-1}=\infty

In order for that to occur, the exponent of x x must be greater than 0 0 .

2 a − 1 > 0 ⟹ a > 1 2 2a-1>0\implies a>\frac{1}{2}

Thus, a > 1 2 a>\frac{1}{2} . Now, if this condition is satisfied,

lim ⁡ x → ∞ π ∫ 1 ∞ x − 2 a d x = − π 1 − 2 a = π 2 a − 1 \lim_{x\to\infty} \pi \int_1^{\infty} x^{-2a} dx=-\frac{\pi}{1-2a}=\frac{\pi}{2a-1}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Didn't they say x > 1, meaning it's safe to include a = 1/2

John Morcos - 2 years, 9 months ago

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Inputting a = 1 2 a=\frac{1}{2} into 1 ( 1 − 2 a ) x 2 a − 1 \frac{1}{(1-2a)x^{2a-1}} gives 1 0 \frac{1}{0} , which is clearly not defined.

Andrei Li - 2 years, 9 months ago

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