To solve the line integral of the curve given by $I = \int_C (x^4 - y^4) \, dl$, where the curve $C$ is defined by $x^2 + y^2 = 4$, we can proceed as follows:

Step 1: Parameterize the Curve

The curve $C$ is a circle of radius 2 centered at the origin. We can parameterize this curve using the standard polar coordinates: $x = 2\cos\theta, \quad y = 2\sin\theta$ where $\theta$ ranges from $0$ to $2\pi$.

Step 2: Express $dl$

The differential length element $dl$ along the curve can be expressed as: $dl = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta$ where: $\frac{dx}{d\theta} = -2\sin\theta, \quad \frac{dy}{d\theta} = 2\cos\theta$ Therefore: $dl = \sqrt{(-2\sin\theta)^2 + (2\cos\theta)^2} \, d\theta = \sqrt{4(\sin^2\theta + \cos^2\theta)} \, d\theta = 2 \, d\theta$

Step 3: Substitute into the Integral

Substitute the parameterization into the integrand $x^4 - y^4$: $x^4 = (2\cos\theta)^4 = 16\cos^4\theta, \quad y^4 = (2\sin\theta)^4 = 16\sin^4\theta$ Thus: $x^4 - y^4 = 16(\cos^4\theta - \sin^4\theta)$

The integral now becomes: $I = \int_0^{2\pi} 16(\cos^4\theta - \sin^4\theta) \times 2 \, d\theta = 32 \int_0^{2\pi} (\cos^4\theta - \sin^4\theta) \, d\theta$

Step 4: Simplify the Integral

Use the identity: $\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)$ Since $\cos^2\theta + \sin^2\theta = 1$, we have: $\cos^4\theta - \sin^4\theta = \cos^2\theta - \sin^2\theta = \cos 2\theta$

Therefore, the integral simplifies to: $I = 32 \int_0^{2\pi} \cos 2\theta \, d\theta$

Step 5: Evaluate the Integral

The integral of $\cos 2\theta$ over one full period $[0, 2\pi]$ is zero because $\cos 2\theta$ is a periodic function that oscillates symmetrically about the x-axis: $\int_0^{2\pi} \cos 2\theta \, d\theta = 0$

Final Answer

Thus, the value of the line integral is: $I = 0$

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. What is the physical interpretation of a line integral?
2. How does parameterization simplify evaluating line integrals?
3. What are the different types of line integrals in vector calculus?
4. Can you explain the difference between a line integral and a surface integral?
5. How do you evaluate a line integral along a non-circular curve?

Tip: Always check the symmetry of the integrand when evaluating integrals over a complete period, as it can simplify the computation significantly.