To evaluate the line integral $\int_C xy \, ds$ where the curve $C$ is given by the parametric equations $x = t^2$ and $y = 2t$ for $0 \le t \le 5$, we will follow these steps:

1. Parameterize the curve: The curve is already parameterized with $x(t) = t^2$ and $y(t) = 2t$.
2. Compute $ds$: $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$.
3. Substitute $x$ and $y$ into the integral: Replace $x$ and $y$ with their parameterized forms.
4. Evaluate the integral.

Let's go through these steps one by one.

Step 1: Parameterize the Curve

The curve is parameterized as: $x(t) = t^2$ $y(t) = 2t$ with $t$ ranging from 0 to 5.

Step 2: Compute $ds$

First, we find $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t$ $\frac{dy}{dt} = \frac{d}{dt}(2t) = 2$

Then, we compute $ds$: $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \sqrt{(2t)^2 + 2^2} \, dt = \sqrt{4t^2 + 4} \, dt = \sqrt{4(t^2 + 1)} \, dt = 2\sqrt{t^2 + 1} \, dt$

Step 3: Substitute $x$ and $y$ into the Integral

We substitute $x = t^2$ and $y = 2t$ into the integral: $\int_C xy \, ds = \int_0^5 (t^2)(2t) \cdot 2\sqrt{t^2 + 1} \, dt = \int_0^5 2t^3 \cdot 2\sqrt{t^2 + 1} \, dt = 4 \int_0^5 t^3 \sqrt{t^2 + 1} \, dt$

Step 4: Evaluate the Integral

To evaluate the integral $4 \int_0^5 t^3 \sqrt{t^2 + 1} \, dt$, we use the substitution method. Let $u = t^2 + 1$. Then $du = 2t \, dt$ or $dt = \frac{du}{2t}$.

When $t = 0$, $u = 1$. When $t = 5$, $u = 26$.

Substitute $u$ and $dt$ into the integral: $4 \int_0^5 t^3 \sqrt{t^2 + 1} \, dt = 4 \int_1^{26} t^3 \sqrt{u} \cdot \frac{du}{2t}$ $= 4 \int_1^{26} \frac{t^2 \sqrt{u}}{2} \, du$

Since $t^2 = u - 1$, we have: $4 \int_1^{26} \frac{(u - 1) \sqrt{u}}{2} \, du$ $= 2 \int_1^{26} (u - 1) \sqrt{u} \, du$ $= 2 \int_1^{26} (u^{3/2} - u^{1/2}) \, du$

Now, integrate term by term: $2 \left[ \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right]_1^{26}$ $= 2 \left( \frac{2}{5} (26)^{5/2} - \frac{2}{3} (26)^{3/2} - \left( \frac{2}{5} (1)^{5/2} - \frac{2}{3} (1)^{3/2} \right) \right)$ $= 2 \left( \frac{2}{5} (26)^{5/2} - \frac{2}{3} (26)^{3/2} - \frac{2}{5} + \frac{2}{3} \right)$

Simplify the constants: $= 2 \left( \frac{2}{5} (26)^{5/2} - \frac{2}{3} (26)^{3/2} + \frac{1}{15} \right)$

Finally, calculate the numerical values: $26^{3/2} = 26 \sqrt{26}$ $26^{5/2} = 26^2 \sqrt{26} = 676 \sqrt{26}$

So, $= 2 \left( \frac{2}{5} \times 676 \sqrt{26} - \frac{2}{3} \times 26 \sqrt{26} + \frac{1}{15} \right)$ $= 2 \left( \frac{1352}{5} \sqrt{26} - \frac{52}{3} \sqrt{26} + \frac{1}{15} \right)$ $= 2 \left( 270.4 \sqrt{26} - 17.33 \sqrt{26} + \frac{1}{15} \right)$ $= 2 \left( 253.07 \sqrt{26} + \frac{1}{15} \right)$

Thus, the evaluated line integral is: $\int_C xy \, ds = 2 \left( 253.07 \sqrt{26} + \frac{1}{15} \right)$

If you have any questions or need further details, please let me know. Here are five more questions you might be interested in:

1. How do you evaluate a line integral for a vector field?
2. What is the divergence theorem and how is it used in vector calculus?
3. Can you explain Green's theorem and its applications?
4. How do you solve surface integrals over a given surface?
5. What are some practical applications of line integrals in physics and engineering?

Tip: When evaluating integrals, always consider if a substitution or change of variables can simplify the computation.