To solve the given problem, let’s go through each question step-by-step:

Given Data

We have a function: $f(x, y) = x^2 + y^2$

Two parametrizations are given for the line segment from $(0, 0)$ to $(2, 2)$:

1. Curve $C_1$: $x = t$, $y = t$, where $0 \leq t \leq 2$.
2. Curve $C_2$: $x = 2t$, $y = 2t$, where $0 \leq t \leq 1$.

We need to calculate the line integrals over $C_1$ and $C_2$, and then compare the results.

Step 1: Calculate the Line Integral over $C_1$

The line integral of $f(x, y)$ along $C_1$ is: $\int_{C_1} f(x, y) \, ds$ where $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$.

1.1 Parametrize $ds$ for $C_1$

For $C_1$:

• $x = t$ and $y = t$.
• Then, $f(x, y) = t^2 + t^2 = 2t^2$.

Calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

• $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = 1$.

Thus: $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt = \sqrt{1^2 + 1^2} \, dt = \sqrt{2} \, dt$

1.2 Set Up and Evaluate the Integral for $C_1$

The integral becomes: $\int_{C_1} f(x, y) \, ds = \int_0^2 2t^2 \cdot \sqrt{2} \, dt = 2\sqrt{2} \int_0^2 t^2 \, dt$

Now, evaluate $\int_0^2 t^2 \, dt$: $\int_0^2 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^2 = \frac{8}{3}$

Thus: $\int_{C_1} f(x, y) \, ds = 2\sqrt{2} \cdot \frac{8}{3} = \frac{16\sqrt{2}}{3}$

Step 2: Calculate the Line Integral over $C_2$

The line integral of $f(x, y)$ along $C_2$ is: $\int_{C_2} f(x, y) \, ds$ where $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt$.

2.1 Parametrize $ds$ for $C_2$

For $C_2$:

• $x = 2t$ and $y = 2t$.
• Then, $f(x, y) = (2t)^2 + (2t)^2 = 4t^2 + 4t^2 = 8t^2$.

Calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

• $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 2$.

Thus: $ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt = \sqrt{2^2 + 2^2} \, dt = \sqrt{8} \, dt = 2\sqrt{2} \, dt$

2.2 Set Up and Evaluate the Integral for $C_2$

The integral becomes: $\int_{C_2} f(x, y) \, ds = \int_0^1 8t^2 \cdot 2\sqrt{2} \, dt = 16\sqrt{2} \int_0^1 t^2 \, dt$

Now, evaluate $\int_0^1 t^2 \, dt$: $\int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3}$

Thus: $\int_{C_2} f(x, y) \, ds = 16\sqrt{2} \cdot \frac{1}{3} = \frac{16\sqrt{2}}{3}$

Step 3: Comparison of Integrals

Both integrals yield the same result: $\int_{C_1} f(x, y) \, ds = \int_{C_2} f(x, y) \, ds = \frac{16\sqrt{2}}{3}$

Summary of Answers

1. The line integral over $C_1$ is $\frac{16\sqrt{2}}{3}$.
2. The line integrals over $C_1$ and $C_2$ are equal, both giving $\frac{16\sqrt{2}}{3}$.

Would you like further details on this solution or have additional questions?

Here are some related questions you might consider:

1. How would the result change if we used a different function $f(x, y)$?
2. What is the geometric interpretation of this line integral?
3. How does the parametrization of a path affect the line integral result?
4. Could we solve this integral using a different coordinate system?
5. How would the calculation differ if we integrated over a different path from $(0, 0)$ to $(2, 2)$?

Tip: When evaluating line integrals, ensure you correctly calculate $ds$ for each parametrization, as it directly affects the integral’s result.