Area and volume Consider the function f(x) = (9 + x²)^(-1/2) and the region R on the interval [0, 4] (see figure).

b. Find the volume of the solid generated when R is revolved about the x-axis.

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Identify the function and the region: The function is given as \(f(x) = (9 + x^{2})^{-1/2}\), and the region \(R\) is bounded by this curve, the x-axis, and the vertical lines \(x=0\) and \(x=4\).

Recall the formula for the volume of a solid of revolution about the x-axis: When a region bounded by \(y=f(x)\) is revolved about the x-axis, the volume \(V\) is given by the integral \(V = \pi \int_{a}^{b} [f(x)]^{2} \, dx\).

Set up the integral for the volume: Substitute \(f(x)\) into the formula, so the volume is \(V = \pi \int_{0}^{4} \left((9 + x^{2})^{-1/2}\right)^{2} \, dx\).

Simplify the integrand: Since squaring \((9 + x^{2})^{-1/2}\) gives \((9 + x^{2})^{-1}\), the integral becomes \(V = \pi \int_{0}^{4} \frac{1}{9 + x^{2}} \, dx\).

Evaluate the integral: Recognize that the integral of \(\frac{1}{a^{2} + x^{2}}\) with respect to \(x\) is \(\frac{1}{a} \arctan\left(\frac{x}{a}\right) + C\). Use this to express the definite integral from 0 to 4, then multiply by \(\pi\) to find the volume.

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