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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

Disk Method

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.

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.

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.