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The given problem asks to evaluate the line integral 

\[
I = \int_C \sqrt{x^2 + y^2} \, ds
\]

where \( C \) is the curve defined by the equation of the circle:

\[
x^2 + y^2 = ax.
\]

### Step-by-Step Solution:

#### 1. Simplify the circle equation:
Rewriting the circle's equation:
\[
x^2 + y^2 = ax
\]
we can rearrange it:
\[
x^2 - ax + y^2 = 0.
\]
Complete the square for \( x^2 - ax \):
\[
x^2 - ax = \left(x - \frac{a}{2}\right)^2 - \frac{a^2}{4}.
\]
Thus, the circle's equation becomes:
\[
\left(x - \frac{a}{2}\right)^2 + y^2 = \frac{a^2}{4}.
\]
This is a circle with:
- Center at \( \left(\frac{a}{2}, 0\right) \),
- Radius \( \frac{a}{2} \).

#### 2. Represent the curve \( C \) parametrically:
Using the parametric representation of a circle:
\[
x = \frac{a}{2} + \frac{a}{2} \cos t, \quad y = \frac{a}{2} \sin t, \quad t \in [0, 2\pi].
\]

#### 3. Arc length element \( ds \):
The arc length element is:
\[
ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt.
\]
From the parameterization:
\[
\frac{dx}{dt} = -\frac{a}{2} \sin t, \quad \frac{dy}{dt} = \frac{a}{2} \cos t,
\]
so:
\[
ds = \sqrt{\left(-\frac{a}{2} \sin t\right)^2 + \left(\frac{a}{2} \cos t\right)^2} \, dt.
\]
\[
ds = \sqrt{\frac{a^2}{4} (\sin^2 t + \cos^2 t)} \, dt = \frac{a}{2} \, dt.
\]

#### 4. Expression for \( \sqrt{x^2 + y^2} \):
Using the parameterization:
\[
x^2 + y^2 = \left(\frac{a}{2} + \frac{a}{2} \cos t\right)^2 + \left(\frac{a}{2} \sin t\right)^2.
\]
Simplify:
\[
x^2 + y^2 = \frac{a^2}{4} + \frac{a^2}{2} \cos t + \frac{a^2}{4} \cos^2 t + \frac{a^2}{4} \sin^2 t.
\]
Using \( \cos^2 t + \sin^2 t = 1 \):
\[
x^2 + y^2 = \frac{a^2}{4} + \frac{a^2}{2} \cos t + \frac{a^2}{4}.
\]
\[
x^2 + y^2 = \frac{a^2}{2} (1 + \cos t).
\]
Thus:
\[
\sqrt{x^2 + y^2} = \sqrt{\frac{a^2}{2} (1 + \cos t)} = \frac{a}{\sqrt{2}} \sqrt{1 + \cos t}.
\]

#### 5. Substitute into the integral:
The integral becomes:
\[
I = \int_C \sqrt{x^2 + y^2} \, ds = \int_0^{2\pi} \frac{a}{\sqrt{2}} \sqrt{1 + \cos t} \cdot \frac{a}{2} \, dt.
\]
\[
I = \frac{a^2}{2\sqrt{2}} \int_0^{2\pi} \sqrt{1 + \cos t} \, dt.
\]

#### 6. Simplify \( \sqrt{1 + \cos t} \):
Using the trigonometric identity:
\[
1 + \cos t = 2 \cos^2 \frac{t}{2},
\]
we have:
\[
\sqrt{1 + \cos t} = \sqrt{2} \cos \frac{t}{2}.
\]
Thus, the integral becomes:
\[
I = \frac{a^2}{2\sqrt{2}} \int_0^{2\pi} \sqrt{2} \cos \frac{t}{2} \, dt.
\]
\[
I = \frac{a^2}{2} \int_0^{2\pi} \cos \frac{t}{2} \, dt.
\]

#### 7. Solve the integral:
The integral of \( \cos \frac{t}{2} \) is:
\[
\int \cos \frac{t}{2} \, dt = 2 \sin \frac{t}{2}.
\]
Evaluate from \( 0 \) to \( 2\pi \):
\[
\int_0^{2\pi} \cos \frac{t}{2} \, dt = 2 \sin \frac{t}{2} \Big|_0^{2\pi} = 2 (\sin \pi - \sin 0) = 2 (0 - 0) = 0.
\]

#### 8. Final Answer:
Since the integral evaluates to zero:
\[
I = 0.
\]

Let me know if you'd like further clarifications or detailed steps!

---

### Related Questions:
1. What is the geometrical significance of this integral?
2. How would the result change if the curve were a semicircle?
3. Can we solve this integral using Green's theorem or symmetry arguments?
4. What happens if the weight \( \sqrt{x^2 + y^2} \) is replaced with \( x \)?
5. How can we parametrize the circle differently to simplify calculations?

### Tip:
When working with circular or elliptical integrals, symmetry can often simplify the computation significantly!

Evaluate the line integral I = ∫_C √(x² + y²) ds, where C is defined by the curve x² + y² = ax.

Learn how to compute the line integral I = ∫_C √(x² + y²) ds where C is the curve defined by x² + y² = ax. This problem involves parameterizing the curve, simplifying trigonometric identities, and calculating the arc length element ds step-by-step. Suitable for undergraduate mathematics students.

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