Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = |x^3|$ and $y = -x^2 + 2$ about the $x$-axis.

We find the intersections of the cubic curve $|x^3|$ with the parabola, then apply the washer method.

Step 1: Find intersection points

For $x\ge0$: $x^3=-x^2+2 \implies x^3+x^2-2=0$. Testing $x=1$: $1+1-2=0$ ✓. Factoring: $(x-1)(x^2+2x+2)=0$; the quadratic has discriminant $4-8=-4<0$ (no real roots), so $x=1$ is the only nonnegative solution.

For $x<0$: $-x^3=-x^2+2 \implies x^3-x^2+2=0$. Testing $x=-1$: $-1-1+2=0$ ✓. So $x=-1$ is a solution (and by the same quadratic-discriminant argument the only one).

The curves meet at $(-1,1)$ and $(1,1)$.

Step 2: Determine which curve is above

At $x=0$: $|x^3|=0$ and $-x^2+2=2$, so the parabola is above $y=|x^3|$ on $[-1,1]$.

Step 3: Set up the washer method

At position $x\in[-1,1]$: outer radius $R(x)=-x^2+2$, inner radius $r(x)=|x^3|$, and $r(x)^2=x^6$:

\[A(x) = \pi\left[(-x^2+2)^2-x^6\right] = \pi(-x^6+x^4-4x^2+4)\]

Step 4: Integrate using symmetry

\[V = 2\int_0^1\pi(-x^6+x^4-4x^2+4)\,dx = 2\pi\left[-\frac{x^7}{7}+\frac{x^5}{5}-\frac{4x^3}{3}+4x\right]_0^1\]

At $x=1$ (common denominator 105): $-\dfrac{15}{105}+\dfrac{21}{105}-\dfrac{140}{105}+\dfrac{420}{105}=\dfrac{286}{105}$.

\[V = 2\pi\cdot\frac{286}{105} = \frac{572\pi}{105}\]

Since $\gcd(572,105)=1$ (Euclidean algorithm confirms), this is already in lowest terms.

\[\boxed{V = \frac{572\pi}{105} \text{ cubic units}}\]

Step 5: Numerical value

\[V = \frac{572\pi}{105} \approx 17.11 \text{ cubic units}\]

Summary

\[\boxed{V = \frac{572\pi}{105} \approx 17.11 \text{ cubic units}}\]

Key insight: Replacing $y=x^2$ (Problem 61's inner boundary) with the steeper $y=|x^3|$ shrinks the inner "hollowed-out" volume, so the total volume ($\frac{572\pi}{105}\approx17.11$) comes out larger than Problem 61's $\frac{76\pi}{15}\approx15.90$, even though both regions share the same outer parabola and the same intersection points.