Find the volume generated by rotating the region bounded

Find the volume generated by rotating the region bounded by the curves $y = x^2$, $x = 0$, $x = 1$, $y = 0$ about: 1° the $x$-axis, 2° the $y$-axis.

We compute each volume of revolution with the method best suited to its axis, then cross-check the $y$-axis result a second way.

Part 1: Rotation about the $x$-axis

Step 1: Disk method

At position $x$, the disk has radius $r(x)=x^2$, so $A(x)=\pi x^4$.

Step 2: Integrate

\[V_x = \pi\int_0^1 x^4\,dx = \pi\left[\frac{x^5}{5}\right]_0^1 = \frac{\pi}{5}\]

\[\boxed{V_x = \frac{\pi}{5} \text{ cubic units}}\]

Part 2: Rotation about the $y$-axis

Step 1: Shell method

A shell at radius $x$ has height $x^2$ and circumference $2\pi x$, so $dV=2\pi x\cdot x^2\,dx = 2\pi x^3\,dx$.

Step 2: Integrate

\[V_y = 2\pi\int_0^1 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^1 = \frac{\pi}{2}\]

\[\boxed{V_y = \frac{\pi}{2} \text{ cubic units}}\]

Step 3: Cross-check via the washer method

Writing $x=\sqrt y$, at height $y\in[0,1]$ the washer has outer radius $1$ and inner radius $\sqrt y$:

\[A(y) = \pi(1-y), \qquad V_y = \pi\int_0^1(1-y)\,dy = \pi\left(1-\frac12\right) = \frac{\pi}{2} \quad\checkmark\]

Comparison

\[\frac{V_y}{V_x} = \frac{\pi/2}{\pi/5} = \frac52 = 2.5\]

numerically, $V_x=\frac\pi5\approx0.628$ and $V_y=\frac\pi2\approx1.571$ cubic units.

Summary

\[\boxed{V_x = \frac{\pi}{5}\approx0.628 \text{ cubic units}, \qquad V_y = \frac{\pi}{2}\approx1.571 \text{ cubic units}}\]

Key insight: The same thin region produces very different solids depending on the axis: a slender paraboloid about the $x$-axis, versus a wider "bowl-shaped" solid about the $y$-axis whose volume is $2.5$ times larger.