The image contains a mathematical problem involving a line integral:

$\int_C (x - y) \, dl,$ where $C$ is defined by the curve $x^2 + y^2 = ax$, with $a > 0$.

Solution:

1. Analyze the curve $x^2 + y^2 = ax$: Rewrite the equation: $x^2 + y^2 - ax = 0.$ This can be expressed as: $x^2 - ax + y^2 = 0,$ completing the square for $x$: $(x - \frac{a}{2})^2 + y^2 = \frac{a^2}{4}.$ Hence, the curve $C$ represents a circle with:

   • Center at $(\frac{a}{2}, 0)$,
   • Radius $\frac{a}{2}$.

2. Parametrize the curve $C$: Using the standard circle parametrization: $x = \frac{a}{2} + \frac{a}{2} \cos\theta, \quad y = \frac{a}{2} \sin\theta, \quad \theta \in [0, 2\pi].$

3. Find $dl$: The differential arc length $dl$ for a parametrized curve is given by: $dl = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta.$ Compute derivatives: $\frac{dx}{d\theta} = -\frac{a}{2} \sin\theta, \quad \frac{dy}{d\theta} = \frac{a}{2} \cos\theta.$ Thus: $dl = \sqrt{\left(-\frac{a}{2} \sin\theta\right)^2 + \left(\frac{a}{2} \cos\theta\right)^2} \, d\theta = \sqrt{\frac{a^2}{4} (\sin^2\theta + \cos^2\theta)} \, d\theta = \frac{a}{2} \, d\theta.$

4. Substitute into the integral: The integrand $x - y$ becomes: $x - y = \left(\frac{a}{2} + \frac{a}{2} \cos\theta\right) - \frac{a}{2} \sin\theta = \frac{a}{2} (1 + \cos\theta - \sin\theta).$ Substitute $x - y$ and $dl$ into the integral: $\int_C (x - y) \, dl = \int_0^{2\pi} \frac{a}{2} (1 + \cos\theta - \sin\theta) \cdot \frac{a}{2} \, d\theta.$ Simplify: $\int_C (x - y) \, dl = \frac{a^2}{4} \int_0^{2\pi} (1 + \cos\theta - \sin\theta) \, d\theta.$

5. Evaluate the integral: Split the integral: $\int_C (x - y) \, dl = \frac{a^2}{4} \left( \int_0^{2\pi} 1 \, d\theta + \int_0^{2\pi} \cos\theta \, d\theta - \int_0^{2\pi} \sin\theta \, d\theta \right).$ Compute each term:

   • $\int_0^{2\pi} 1 \, d\theta = 2\pi$,
   • $\int_0^{2\pi} \cos\theta \, d\theta = 0$ (over a full period),
   • $\int_0^{2\pi} \sin\theta \, d\theta = 0$ (over a full period).

   Thus: $\int_C (x - y) \, dl = \frac{a^2}{4} \cdot 2\pi = \frac{\pi a^2}{2}.$

Final Answer:

$\int_C (x - y) \, dl = \frac{\pi a^2}{2}.$

Let me know if you'd like further details or have questions! Here are 5 related problems to explore further:

1. What is the general formula for a line integral involving a scalar field?
2. How does one parametrize other conic sections, like ellipses or parabolas, for line integrals?
3. How do you compute $\int_C \mathbf{F} \cdot \mathbf{T} \, dl$, where $\mathbf{T}$ is the tangent vector?
4. What is the geometric interpretation of a line integral over a closed curve?
5. How would the integral change if the curve $C$ had a different orientation?

Tip: Always verify whether a curve is closed and simple before applying certain theorems like Green's or Stokes'!