 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

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

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

### Step 2: Compute the Arc Length Element \( ds \)
The arc length element \( ds \) is given by:
\[
ds = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta
\]
First, compute the derivatives of \( x(\theta) \) and \( y(\theta) \):
\[
\frac{dx}{d\theta} = -\sin(\theta), \quad \frac{dy}{d\theta} = \cos(\theta)
\]
Now, substitute into the expression for \( ds \):
\[
ds = \sqrt{(-\sin(\theta))^2 + (\cos(\theta))^2} \, d\theta = \sqrt{\sin^2(\theta) + \cos^2(\theta)} \, d\theta = d\theta
\]
So, \( ds = d\theta \).

### Step 3: Set Up the Line Integral
The line integral becomes:
\[
\int_C (9x - 8y) \, ds = \int_0^{\frac{\pi}{2}} \left[9(4 + \cos(\theta)) - 8(4 + \sin(\theta))\right] d\theta
\]
Simplifying the integrand:
\[
9(4 + \cos(\theta)) - 8(4 + \sin(\theta)) = 36 + 9\cos(\theta) - 32 - 8\sin(\theta) = 4 + 9\cos(\theta) - 8\sin(\theta)
\]
Thus, the integral becomes:
\[
\int_0^{\frac{\pi}{2}} \left(4 + 9\cos(\theta) - 8\sin(\theta)\right) d\theta
\]

### Step 4: Evaluate the Integral
We can now evaluate the integral term by term:
\[
\int_0^{\frac{\pi}{2}} 4 \, d\theta = 4\left[\theta\right]_0^{\frac{\pi}{2}} = 4 \times \frac{\pi}{2} = 2\pi
\]
\[
\int_0^{\frac{\pi}{2}} 9\cos(\theta) \, d\theta = 9\left[\sin(\theta)\right]_0^{\frac{\pi}{2}} = 9 \times (1 - 0) = 9
\]
\[
\int_0^{\frac{\pi}{2}} -8\sin(\theta) \, d\theta = -8\left[-\cos(\theta)\right]_0^{\frac{\pi}{2}} = -8 \times (-1 - (-0)) = 8
\]

### Step 5: Final Answer
Now, sum the results of the integrals:
\[
2\pi + 9 + 8 = 2\pi + 17
\]

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

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.

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(9x−8y)ds.

Learn how to compute the line integral ∫C(9x−8y)ds over the arc of a circle in the first quadrant. This calculus problem covers parametric equations, the arc length formula, and trigonometric integrals, providing a step-by-step solution for advanced students. Suitable for undergraduate calculus students.

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