Find the volume generated by rotating the region bounded

Find the volume generated by rotating the region bounded by the curve $g(x) = \int_0^x \frac{f(t^2)}{t^2} \, dt$, where $f(x) = \frac{x^{3/2}}{1 + x^2}$, and the lines $x = 0$, $x = 1$, $y = 0$ about the $x$-axis.

The key to this problem is recognizing that $g(x)$ itself has an elementary closed form — a simplification the most tempting "just integrate numerically" approach misses entirely.

Step 1: Simplify the integrand of $g$

With $f(x)=\dfrac{x^{3/2}}{1+x^2}$:

\[f(t^2) = \frac{(t^2)^{3/2}}{1+(t^2)^2} = \frac{t^3}{1+t^4}\]

\[\frac{f(t^2)}{t^2} = \frac{t^3}{t^2(1+t^4)} = \frac{t}{1+t^4}\]

So $g(x) = \displaystyle\int_0^x \frac{t}{1+t^4}\,dt$.

Step 2: Find a closed form for $g(x)$

Let $u=t^2$, $du=2t\,dt$. When $t=0$: $u=0$; when $t=x$: $u=x^2$.

\[g(x) = \int_0^{x^2} \frac{1}{1+u^2}\cdot\frac12\,du = \frac12\left[\arctan u\right]_0^{x^2} = \frac12\arctan(x^2)\]

This is the crucial simplification: $g(x)=\dfrac12\arctan(x^2)$ is completely elementary, so the "nested integral" in the problem statement was never actually necessary to keep nested.

Step 3: Set up the volume integral

Using the disk method about the $x$-axis:

\[V = \pi\int_0^1 [g(x)]^2\,dx = \pi\int_0^1 \left[\frac12\arctan(x^2)\right]^2 dx = \frac{\pi}{4}\int_0^1 \arctan^2(x^2)\,dx\]

Step 4: Evaluate

The remaining integral $\int_0^1\arctan^2(x^2)\,dx$ does not reduce to elementary functions (it does not correspond to any standard closed form in terms of $\pi$, $\ln$, or $\arctan$ alone), so we evaluate it numerically to high precision:

\[\int_0^1 \arctan^2(x^2)\,dx \approx 0.15045\]

\[V = \frac{\pi}{4}\times0.15045 \approx 0.1182\]

\[\boxed{V \approx 0.1182 \text{ cubic units}}\]

Step 5: Sanity check at the endpoint

At $x=1$: $g(1)=\dfrac12\arctan(1)=\dfrac{\pi}{8}\approx0.3927$, so the radius never exceeds about $0.393$ over $[0,1]$ — consistent with a volume well under $\pi\cdot(0.393)^2\cdot1\approx0.485$ (an upper bound from a cylinder of that radius), which our computed value comfortably satisfies.

Summary

\[\boxed{g(x) = \frac12\arctan(x^2), \qquad V = \frac{\pi}{4}\int_0^1\arctan^2(x^2)\,dx \approx 0.1182 \text{ cubic units}}\]

Key insight: Although the volume integral itself must be finished numerically, the substitution $u=t^2$ collapses the defining integral for $g(x)$ down to the single elementary expression $\frac12\arctan(x^2)$ — the genuinely hard part of the problem is spotting this simplification, not evaluating the integral once you have it.