Find the volume generated by rotating the region bounded

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

We locate the intersections, confirm which curve is outer, and apply the washer method using symmetry.

Step 1: Find intersection points

For $x\ge0$: $-x^2+2=x \implies x^2+x-2=0 \implies (x+2)(x-1)=0$, so $x=1$.

For $x < 0$: $-x^2+2=-x \implies x^2-x-2=0 \implies (x-2)(x+1)=0$, so $x=-1$.

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

Step 2: Determine which curve is above

At $x=0$: parabola gives $y=2$; $|x|$ gives $y=0$. So the parabola lies above $y=|x|$ 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 $r(x)=|x|$.

\[A(x) = \pi\left[(-x^2+2)^2-x^2\right] = \pi\left[(x^4-4x^2+4)-x^2\right] = \pi(x^4-5x^2+4)\]

Step 4: Integrate using symmetry

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

At $x=1$: $\dfrac15-\dfrac53+4 = \dfrac{3-25+60}{15}=\dfrac{38}{15}$.

\[V = 2\pi\cdot\frac{38}{15} = \frac{76\pi}{15}\]

\[\boxed{V = \frac{76\pi}{15} \text{ cubic units}}\]

Step 5: Numerical value

\[V = \frac{76\pi}{15} \approx 15.90 \text{ cubic units}\]

Summary

\[\boxed{V = \frac{76\pi}{15} \approx 15.90 \text{ cubic units}}\]

Key insight: The V-shaped curve $y=|x|$ acts as an inner boundary that "hollows out" the solid generated by the parabola alone, and the washer method handles the resulting lens-shaped cross-section cleanly despite the corner in $|x|$ at the origin.