Problem 34
Question
Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=\frac{\sqrt{x}}{\sqrt[3]{x^{2}+1}}\) and the \(x\) -axis on the interval \([0, \infty)\) is revolved about the \(x\) -axis.
Step-by-Step Solution
Verified Answer
Answer: No, the volume of the solid of revolution does not exist, as the integral representing the volume diverges.
1Step 1: Set up the integral
For \(f(x) = \frac{\sqrt{x}}{\sqrt[3]{x^2+1}}\) on \([0, \infty)\), the volume of revolution about the x-axis is \(V = \pi\int_0^\infty \frac{x}{(x^2+1)^{2/3}}\,dx\).
2Step 2: Check convergence
For large \(x\): \(\frac{x}{(x^2+1)^{2/3}} \approx \frac{x}{x^{4/3}} = x^{-1/3}\), which is not integrable on \([1,\infty)\). The integral diverges, so the volume does not exist.
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