Math Problem Statement

If C is the part of the circle (x4)2+(y4)2=1 in the first quadrant, find the following line integral with respect to arc length.

∫C(9x−8y)ds

Solution

We are asked to find the line integral of C(9x8y)ds\int_C (9x - 8y) \, ds, where CC is the part of the circle defined by the equation (x4)2+(y4)2=1(x - 4)^2 + (y - 4)^2 = 1 that lies in the first quadrant. Here, dsds represents the differential arc length, and the circle is centered at (4,4)(4, 4) with a radius of 1.

Step 1: Parametrize the Circle

The equation of the circle can be written as: (x4)2+(y4)2=1(x - 4)^2 + (y - 4)^2 = 1 We can parametrize this circle in terms of an angle θ\theta, where θ[0,π2]\theta \in [0, \frac{\pi}{2}] since we are only interested in the first quadrant: x(θ)=4+cos(θ),y(θ)=4+sin(θ)x(\theta) = 4 + \cos(\theta), \quad y(\theta) = 4 + \sin(\theta) where θ[0,π2]\theta \in [0, \frac{\pi}{2}].

Step 2: Compute the Arc Length Element dsds

The arc length element 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 First, compute the derivatives of x(θ)x(\theta) and y(θ)y(\theta): dxdθ=sin(θ),dydθ=cos(θ)\frac{dx}{d\theta} = -\sin(\theta), \quad \frac{dy}{d\theta} = \cos(\theta) Now, substitute into the expression for dsds: ds=(sin(θ))2+(cos(θ))2dθ=sin2(θ)+cos2(θ)dθ=dθds = \sqrt{(-\sin(\theta))^2 + (\cos(\theta))^2} \, d\theta = \sqrt{\sin^2(\theta) + \cos^2(\theta)} \, d\theta = d\theta So, ds=dθds = d\theta.

Step 3: Set Up the Line Integral

The line integral becomes: C(9x8y)ds=0π2[9(4+cos(θ))8(4+sin(θ))]dθ\int_C (9x - 8y) \, ds = \int_0^{\frac{\pi}{2}} \left[9(4 + \cos(\theta)) - 8(4 + \sin(\theta))\right] d\theta Simplifying the integrand: 9(4+cos(θ))8(4+sin(θ))=36+9cos(θ)328sin(θ)=4+9cos(θ)8sin(θ)9(4 + \cos(\theta)) - 8(4 + \sin(\theta)) = 36 + 9\cos(\theta) - 32 - 8\sin(\theta) = 4 + 9\cos(\theta) - 8\sin(\theta) Thus, the integral becomes: 0π2(4+9cos(θ)8sin(θ))dθ\int_0^{\frac{\pi}{2}} \left(4 + 9\cos(\theta) - 8\sin(\theta)\right) d\theta

Step 4: Evaluate the Integral

We can now evaluate the integral term by term: 0π24dθ=4[θ]0π2=4×π2=2π\int_0^{\frac{\pi}{2}} 4 \, d\theta = 4\left[\theta\right]_0^{\frac{\pi}{2}} = 4 \times \frac{\pi}{2} = 2\pi 0π29cos(θ)dθ=9[sin(θ)]0π2=9×(10)=9\int_0^{\frac{\pi}{2}} 9\cos(\theta) \, d\theta = 9\left[\sin(\theta)\right]_0^{\frac{\pi}{2}} = 9 \times (1 - 0) = 9 0π28sin(θ)dθ=8[cos(θ)]0π2=8×(1(0))=8\int_0^{\frac{\pi}{2}} -8\sin(\theta) \, d\theta = -8\left[-\cos(\theta)\right]_0^{\frac{\pi}{2}} = -8 \times (-1 - (-0)) = 8

Step 5: Final Answer

Now, sum the results of the integrals: 2π+9+8=2π+172\pi + 9 + 8 = 2\pi + 17

Thus, the value of the line integral is: 2π+17\boxed{2\pi + 17}

Would you like more details or explanations about any step?

Here are 5 related questions you might explore:

  1. How would the result change if the circle were centered at a different point?
  2. Can the same approach be used for ellipses, and how?
  3. What is the significance of the line integral in physical contexts?
  4. How does the parametrization change for different sections of the circle?
  5. Can this integral be solved in polar coordinates directly?

Tip: When solving line integrals, always ensure that the limits of the parameterization correctly represent the section of the curve being integrated over.

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Math Problem Analysis

Mathematical Concepts

Line Integrals
Parametric Equations
Calculus
Arc Length
Trigonometry

Formulas

Equation of a circle: (x - h)^2 + (y - k)^2 = r^2
Parametric equations: x(θ) = h + cos(θ), y(θ) = k + sin(θ)
Arc length element: ds = sqrt((dx/dθ)^2 + (dy/dθ)^2) dθ
Line integral: ∫C f(x, y) ds

Theorems

Pythagorean Theorem
Fundamental Theorem of Line Integrals

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 and above)