1. **Problem Statement:** We have two functions $f(x) = x^2 - 6$ and $g(x) = x^2 - 5$ defined on the interval $0 \leq x \leq 1$. We want to find the volume of the solid formed by rotating the area between these curves about the $x$-axis. 2. **Understanding the curves:** Both functions are parabolas shifted vertically. Since $g(x) = x^2 - 5$ is always 1 unit above $f(x) = x^2 - 6$ on the interval, the area between them is a thin horizontal strip of height 1. 3. **Formula for volume of solid of revolution:** When rotating around the $x$-axis, the volume $V$ of the solid formed by the area between two curves $y = R(x)$ (outer radius) and $y = r(x)$ (inner radius) from $a$ to $b$ is: $$ V = \pi \int_a^b \left(R(x)^2 - r(x)^2\right) dx $$ 4. **Identify radii:** Here, the outer radius is the distance from the $x$-axis to the upper curve $g(x)$, and the inner radius is the distance to the lower curve $f(x)$. Since both are below the $x$-axis (negative values), the radius is the absolute value of the function: $$ R(x) = |f(x)| = 6 - x^2, \quad r(x) = |g(x)| = 5 - x^2 $$ 5. **Set up the integral:** $$ V = \pi \int_0^1 \left((6 - x^2)^2 - (5 - x^2)^2\right) dx $$ 6. **Expand the squares:** $$ (6 - x^2)^2 = 36 - 12x^2 + x^4 $$ $$ (5 - x^2)^2 = 25 - 10x^2 + x^4 $$ 7. **Subtract inside the integral:** $$ (36 - 12x^2 + x^4) - (25 - 10x^2 + x^4) = (36 - 25) + (-12x^2 + 10x^2) + (x^4 - x^4) = 11 - 2x^2 $$ 8. **Integral becomes:** $$ V = \pi \int_0^1 (11 - 2x^2) dx $$ 9. **Integrate term-by-term:** $$ \int_0^1 11 dx = 11x \Big|_0^1 = 11 $$ $$ \int_0^1 2x^2 dx = 2 \cdot \frac{x^3}{3} \Big|_0^1 = \frac{2}{3} $$ 10. **Calculate volume:** $$ V = \pi \left(11 - \frac{2}{3}\right) = \pi \cdot \frac{33 - 2}{3} = \frac{31\pi}{3} $$ **Final answer:** $$ \boxed{V = \frac{31\pi}{3}} $$ This is the volume of the solid formed by rotating the area between the two parabolas about the $x$-axis over $[0,1]$.