Math Problem Statement

If C is the part of the circle (x/4)^2 + (y/4)^2 = 1 in the first quadrant, find the following line integral with respect to arc length: ∫_C (2x - 9y) ds.

Solution

The problem involves evaluating a line integral over the curve CC, which is the part of the circle

(x4)2+(y4)2=1\left( \frac{x}{4} \right)^2 + \left( \frac{y}{4} \right)^2 = 1

in the first quadrant. The line integral to evaluate is:

C(2x9y)ds\int_C (2x - 9y) \, ds

where dsds represents the differential arc length.

Step-by-Step Solution:

  1. Parameterization of the Curve: The given equation is a circle of radius 4 centered at the origin, but it's restricted to the first quadrant. We can parameterize it using:

    x=4cosθ,y=4sinθx = 4 \cos \theta, \quad y = 4 \sin \theta with θ\theta ranging from 00 to π2\frac{\pi}{2}.

  2. Differential Arc Length dsds: The differential arc length dsds is given by:

    ds=(dxdθ)2+(dydθ)2dθds = \sqrt{\left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2} \, d\theta

    Substituting x=4cosθx = 4 \cos \theta and y=4sinθy = 4 \sin \theta:

    dxdθ=4sinθ,dydθ=4cosθ\frac{dx}{d\theta} = -4 \sin \theta, \quad \frac{dy}{d\theta} = 4 \cos \theta

    ds=(4sinθ)2+(4cosθ)2dθ=16sin2θ+16cos2θdθ=4dθds = \sqrt{(-4 \sin \theta)^2 + (4 \cos \theta)^2} \, d\theta = \sqrt{16 \sin^2 \theta + 16 \cos^2 \theta} \, d\theta = 4 \, d\theta

  3. Substitute in the Line Integral: Substituting x=4cosθx = 4 \cos \theta, y=4sinθy = 4 \sin \theta, and ds=4dθds = 4 \, d\theta into the integral:

    C(2x9y)ds=0π/2[2(4cosθ)9(4sinθ)](4dθ)\int_C (2x - 9y) \, ds = \int_0^{\pi/2} \left[ 2(4 \cos \theta) - 9(4 \sin \theta) \right] (4 \, d\theta)

    Simplifying:

    0π/2[8cosθ36sinθ](4dθ)\int_0^{\pi/2} \left[ 8 \cos \theta - 36 \sin \theta \right] (4 \, d\theta)

    0π/2(32cosθ144sinθ)dθ\int_0^{\pi/2} (32 \cos \theta - 144 \sin \theta) \, d\theta

  4. Evaluating the Integral:

    0π/232cosθdθ0π/2144sinθdθ\int_0^{\pi/2} 32 \cos \theta \, d\theta - \int_0^{\pi/2} 144 \sin \theta \, d\theta

    The integrals of cosθ\cos \theta and sinθ\sin \theta are:

    cosθdθ=sinθandsinθdθ=cosθ\int \cos \theta \, d\theta = \sin \theta \quad \text{and} \quad \int \sin \theta \, d\theta = -\cos \theta

    Therefore:

    [ 32 \left[ \sin \theta \right]_0^{\pi/2} - 144 \left[ -\cos \theta \right]_0^{\pi/2} ]

    32(10)144(0(1))32(1 - 0) - 144(0 - (-1))

    32+144=17632 + 144 = 176

Thus, the value of the line integral is 176176.

Would you like any additional details or have any questions?

Here are some related questions:

  1. How do we parameterize different curves for line integrals?
  2. What is the significance of the differential arc length dsds?
  3. How do we handle line integrals over more complex curves or surfaces?
  4. Can line integrals be used to calculate physical quantities like work done?
  5. What is the relationship between line integrals and Green's Theorem?

Tip: Always double-check the parameterization and limits of integration when working with line integrals, as they significantly affect the result.

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Math Problem Analysis

Mathematical Concepts

Line Integrals
Parametrization
Arc Length

Formulas

Parametrization: x = 4 cos(θ), y = 4 sin(θ)
Arc Length: ds = sqrt((dx/dθ)^2 + (dy/dθ)^2) dθ
Line Integral: ∫_C (2x - 9y) ds

Theorems

Trigonometric Identities
Integral Evaluation

Suitable Grade Level

Undergraduate - Calculus II