The problem involves calculating the line integral of $7y\cos(2x) \, dx + 2xy \, dy$ along the closed curve $C$, which consists of three line segments:

1. From $(0, 0)$ to $(4, 6)$
2. From $(4, 6)$ to $(0, 6)$
3. From $(0, 6)$ to $(0, 0)$.

We'll compute the line integral along each segment and sum the results. Here's the step-by-step solution:

Segment 1: From (0, 0) to (4, 6)

The parametric equations for this line segment are:
$x = t, \quad y = \frac{3}{2}t, \quad \text{where } 0 \leq t \leq 4$ Differentials:
$dx = dt, \quad dy = \frac{3}{2} \, dt$

Substituting these into the integral: $\int_C \left( 7y\cos(2x) \, dx + 2xy \, dy \right) = \int_0^4 \left[ 7 \cdot \frac{3}{2}t \cos(2t) \cdot 1 + 2 \cdot t \cdot \frac{3}{2}t \cdot \frac{3}{2} \right] dt$

Simplify: $\int_0^4 \left[ \frac{21}{2}t\cos(2t) + \frac{9}{2}t^2 \right] dt$

Separate the terms: $\frac{21}{2} \int_0^4 t \cos(2t) \, dt + \frac{9}{2} \int_0^4 t^2 \, dt$

1. For $\int_0^4 t\cos(2t) \, dt$, use integration by parts ($u = t$, $dv = \cos(2t) dt$): $\int t \cos(2t) \, dt = \frac{t \sin(2t)}{2} - \frac{\cos(2t)}{4} \Big|_0^4$ Evaluate: [ = \left[ \frac{4 \sin(8)}{2} - \frac{\cos(8)}{4} \right] - \left[ 0 - \frac{\cos(0)}{4} \right] = 2\sin(8) - \frac{\cos(8) + 1}{4} ]

2. For $\int_0^4 t^2 \, dt$: $\int_0^4 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^4 = \frac{64}{3}$

Combine results: $\int_{\text{Segment 1}} = \frac{21}{2} \left( 2\sin(8) - \frac{\cos(8) + 1}{4} \right) + \frac{9}{2} \cdot \frac{64}{3}$

Segment 2: From (4, 6) to (0, 6)

For this horizontal line segment:
$y = 6$, $x$ varies from 4 to 0. Thus:
$dx = -dt, \quad dy = 0$

Substitute into the integral: $\int_C \left( 7y\cos(2x) \, dx + 2xy \, dy \right) = \int_4^0 7 \cdot 6 \cos(2x) (-1) \, dx = -42 \int_4^0 \cos(2x) \, dx$

Change limits to simplify: $= 42 \int_0^4 \cos(2x) \, dx$

Evaluate $\int \cos(2x) dx$: $\int \cos(2x) dx = \frac{\sin(2x)}{2}$

So: $\int_0^4 \cos(2x) \, dx = \frac{\sin(8)}{2} - 0$

Thus: $\int_{\text{Segment 2}} = 42 \cdot \frac{\sin(8)}{2} = 21\sin(8)$

Segment 3: From (0, 6) to (0, 0)

For this vertical line segment:
$x = 0$, $y$ varies from 6 to 0. Thus:
$dx = 0, \quad dy = -dt$

Substitute into the integral: $\int_C \left( 7y\cos(2x) \, dx + 2xy \, dy \right) = \int_6^0 2 \cdot 0 \cdot y (-1) \, dy = 0$

Final Answer

Combine the results from all three segments: $\int_C \left( 7y\cos(2x) \, dx + 2xy \, dy \right) = \int_{\text{Segment 1}} + \int_{\text{Segment 2}} + \int_{\text{Segment 3}}$

Substitute values: $\int_C = \frac{21}{2} \left( 2\sin(8) - \frac{\cos(8) + 1}{4} \right) + \frac{9}{2} \cdot \frac{64}{3} + 21\sin(8)$

Simplify for the final expression. Let me know if you'd like this simplified further!

Questions for Exploration:

1. What are the advantages of using Green's Theorem to compute this integral instead of evaluating along each segment?
2. How do parametric equations simplify line integrals on straight-line segments?
3. Can we always switch limits of integration to simplify computation? Why?
4. What are some real-world applications of line integrals like this?
5. How would the integral change if the curve $C$ was a circle?

Tip:

When applying Green's Theorem, ensure the curve is positively oriented (counterclockwise) to match the theorem's conditions.