line integral of a curve  I=x^4-y^4 dl where the curve is x^2+y^2=4

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.

Learn how to calculate the line integral of x^4 - y^4 dl over the curve defined by x^2 + y^2 = 4. Follow a detailed step-by-step solution using parameterization and trigonometric identities. Suitable for advanced undergraduate students and professionals.

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