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The problem involves evaluating a line integral over the curve \( C \), which is the part of the circle 

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

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

\[
\int_C (2x - 9y) \, ds
\]

where \( ds \) 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:
   
   \[
   x = 4 \cos \theta, \quad y = 4 \sin \theta
   \]
   with \(\theta\) ranging from \(0\) to \(\frac{\pi}{2}\).

2. **Differential Arc Length \( ds \):**
   The differential arc length \( ds \) is given by:

   \[
   ds = \sqrt{\left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2} \, d\theta
   \]

   Substituting \( x = 4 \cos \theta \) and \( y = 4 \sin \theta \):

   \[
   \frac{dx}{d\theta} = -4 \sin \theta, \quad \frac{dy}{d\theta} = 4 \cos \theta
   \]

   \[
   ds = \sqrt{(-4 \sin \theta)^2 + (4 \cos \theta)^2} \, d\theta = \sqrt{16 \sin^2 \theta + 16 \cos^2 \theta} \, d\theta = 4 \, d\theta
   \]

3. **Substitute in the Line Integral:**
   Substituting \( x = 4 \cos \theta \), \( y = 4 \sin \theta \), and \( ds = 4 \, d\theta \) into the integral:

   \[
   \int_C (2x - 9y) \, ds = \int_0^{\pi/2} \left[ 2(4 \cos \theta) - 9(4 \sin \theta) \right] (4 \, d\theta)
   \]

   Simplifying:

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

   \[
   \int_0^{\pi/2} (32 \cos \theta - 144 \sin \theta) \, d\theta
   \]

4. **Evaluating the Integral:**

   \[
   \int_0^{\pi/2} 32 \cos \theta \, d\theta - \int_0^{\pi/2} 144 \sin \theta \, d\theta
   \]

   The integrals of \(\cos \theta\) and \(\sin \theta\) are:

   \[
   \int \cos \theta \, d\theta = \sin \theta \quad \text{and} \quad \int \sin \theta \, d\theta = -\cos \theta
   \]

   Therefore:

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

   \[
   32(1 - 0) - 144(0 - (-1))
   \]

   \[
   32 + 144 = 176
   \]

Thus, the value of the line integral is \( 176 \).

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 \( ds \)?
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.

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.

Evaluate the line integral ∫_C (2x - 9y) ds over the part of the circle (x/4)^2 + (y/4)^2 = 1 in the first quadrant. Learn how to parameterize the curve, compute arc length, and solve the integral step by step.

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