Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].

c. Find the volume of the solid generated when R is revolved about the x-axis.

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Identify the region R bounded by the curve \(f(x) = \frac{1}{3}x^{3}\) and the x-axis on the interval \([0, 2]\). Since \(f(x) \geq 0\) on this interval, the region lies above the x-axis.

Recall the formula for the volume of a solid of revolution generated by revolving a region around the x-axis: \(V = \pi \int_{a}^{b} [f(x)]^{2} \, dx\).

Substitute the given function and interval into the formula: \(V = \pi \int_{0}^{2} \left( \frac{1}{3}x^{3} \right)^{2} \, dx\).

Simplify the integrand: \(\left( \frac{1}{3}x^{3} \right)^{2} = \frac{1}{9} x^{6}\), so the integral becomes \(V = \pi \int_{0}^{2} \frac{1}{9} x^{6} \, dx\).

Set up the integral for evaluation: \(V = \frac{\pi}{9} \int_{0}^{2} x^{6} \, dx\). The next step would be to compute this integral and then multiply by \(\frac{\pi}{9}\) to find the volume.

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Volume of Solids of Revolution

This concept involves finding the volume of a 3D solid formed by rotating a 2D region around an axis. The volume can be computed using integral calculus, often applying the disk or washer method, which sums up infinitesimal circular cross-sections perpendicular to the axis of rotation.

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The disk method calculates volume by slicing the solid into thin disks perpendicular to the axis of rotation. Each disk's volume is approximated by π times the square of the radius (distance from the axis to the function) times the thickness (dx). Integrating these volumes over the interval gives the total volume.

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Definite Integration

Definite integration is used to sum continuous quantities over an interval. In this problem, it allows the calculation of the total volume by integrating the area of cross-sectional disks from x = 0 to x = 2, providing an exact value for the volume of the solid.

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