Math Problem Statement

Find the area of the surface generated when the given curve y = (x^3)/6 + 1/(2x), for 1/2 ≤ x ≤ 2 is revolved about the x-axis.

Solution

The problem asks for the area of the surface generated when the given curve is revolved about the x-axis. The curve provided is:

y=x36+12x,for 12x2y = \frac{x^3}{6} + \frac{1}{2x}, \quad \text{for } \frac{1}{2} \leq x \leq 2

Formula for Surface Area of Revolution

To find the surface area of a curve y=f(x)y = f(x) revolved around the x-axis, we use the surface area formula:

A=2πaby1+(dydx)2dxA = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx

Step-by-Step Process

  1. Differentiate yy with respect to xx: We need to find dydx\frac{dy}{dx}.

    dydx=ddx(x36+12x)\frac{dy}{dx} = \frac{d}{dx} \left(\frac{x^3}{6} + \frac{1}{2x}\right)

    This derivative can be computed as:

    dydx=163x2121x2=x2212x2\frac{dy}{dx} = \frac{1}{6} \cdot 3x^2 - \frac{1}{2} \cdot \frac{1}{x^2} = \frac{x^2}{2} - \frac{1}{2x^2}

  2. Square the derivative: Now, square the result of dydx\frac{dy}{dx}.

    (dydx)2=(x2212x2)2\left(\frac{dy}{dx}\right)^2 = \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2

  3. Set up the surface area integral:

    The formula becomes:

    A=2π122(x36+12x)1+(x2212x2)2dxA = 2\pi \int_{\frac{1}{2}}^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right) \sqrt{1 + \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2} \, dx

  4. Evaluate the integral: This involves solving the integral, which might require symbolic integration tools for the square root term. Let me compute the exact surface area value for you.

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Math Problem Analysis

Mathematical Concepts

Calculus
Surface Area of Revolution
Differentiation
Integration

Formulas

Surface area of revolution: A = 2π ∫ y sqrt(1 + (dy/dx)^2) dx

Theorems

Surface Area of Revolution

Suitable Grade Level

College-level Calculus