Question

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=\left(x^{2}-1\right)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(y\) -axis.

Short answer

Answer: The volume of the solid of revolution is \(V = \frac{2\pi}{3}\).

Step-by-step solution

Set up the integral for the sum of the disk volumes

Since we're revolving around the y-axis, we will use the disk method formula for volumes: $$V = \pi \int_{a}^{b} [r(x)]^2 dx$$ where V is the volume, a and b are the limits of integration, and r(x) is the radius of each disk. In our case, the radius of each disk is given by the x-coordinate, so \(r(x) = x\). The function \(f(x)\) represents the upper boundary of the region, so we have the interval (1,2] for the x-coordinates. The integral we need to compute is $$V = \pi \int_{1}^{2} [x]^2 [(x^2-1)^{-\frac{1}{4}}] dx$$.

Simplify the integral

Simplify the integrand by multiplying the constants and the functions: $$V = \pi \int_{1}^{2} x^2 (x^2-1)^{-\frac{1}{4}} dx$$

Perform integration

To integrate this function, we will use substitution. Let \(u = x^2 - 1\). Then, we have \(du = 2x dx\). So, we will need an extra \(\frac{1}{2}\) factor when we substitute. $$V = \frac{\pi}{2} \int_{0}^{1} u^{-\frac{1}{4}} du$$ Now, we can easily integrate: $$V = \frac{\pi}{2} \left[\frac{4}{3} u^{\frac{3}{4}} \right]_{0}^{1} $$

Evaluate the integral

Plugging in the limits, we find the volume of the solid of revolution: $$ V = \frac{\pi}{2} \left[\frac{4}{3}(1)^{\frac{3}{4}} - \frac{4}{3}(0)^{\frac{3}{4}} \right] = \frac{2\pi}{3}$$ So, the volume of the solid of revolution formed when the region bounded by \(f(x) = (x^2 - 1)^{-\frac{1}{4}}\) and the x-axis on the interval (1,2] is revolved around the y-axis is \(V = \frac{2\pi}{3}\).