Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

문제 6.6.9

6장, 문제 6.6.9

Find the area of the surface generated when the given curve is revolved about the given axis.

y=8√x, for 9≤x≤20; about the x-axis

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Step 1: Recall the formula for the surface area of a curve revolved about the x-axis. The formula is:

S = 2π ∫

_a^b y √(1 + (dy/dx)²) dx, where y is the function being revolved, and dy/dx is its derivative.

Step 2: Identify the given function y = 8√x and compute its derivative dy/dx. Using the power rule,

dy/dx = 8 * (1/2) * x

^-1/2 = 4x^-1/2.

Step 3: Substitute y = 8√x and dy/dx = 4x^-1/2 into the surface area formula. This gives:

S = 2π ∫

₉²⁰ (8√x) √(1 + (4x^-1/2)²) dx.

Step 4: Simplify the expression inside the square root. Compute

(4x

^-1/2)² = 16x^-1, so the integrand becomes:

S = 2π ∫

₉²⁰ (8√x) √(1 + 16x^-1) dx.

Step 5: Set up the integral for evaluation. The integral is now ready to be computed either analytically or numerically:

S = 2π ∫

₉²⁰ (8√x) √(1 + 16/x) dx. Break this into manageable parts or use numerical methods to evaluate the integral.

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주요 개념

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Surface Area of Revolution

The surface area of revolution is calculated by rotating a curve around an axis. The formula involves integrating the length of the curve multiplied by the radius of rotation. For a curve y=f(x) revolved around the x-axis, the surface area S can be expressed as S = 2π ∫[a to b] f(x) √(1 + (f'(x))^2) dx, where f'(x) is the derivative of f(x).

Parametric Representation

In some cases, curves can be represented parametrically, which involves expressing the coordinates of points on the curve as functions of a parameter. For the curve y=8√x, it can be useful to express x in terms of a parameter t, where t=x, to facilitate integration. This representation helps in calculating derivatives and applying the surface area formula effectively.

Definite Integral

A definite integral calculates the accumulation of quantities, such as area, over a specific interval. In this context, the limits of integration (9 to 20) define the section of the curve being revolved. Understanding how to evaluate definite integrals is crucial for finding the total surface area generated by the revolution of the curve around the x-axis.

Find the area of the surface generated when the given curve is revolved about the given axis.y=8√x, for 9≤x≤20; about the x-axis