To solve the line integral $\int_C 3x \, dy$, where $C$ is a positively oriented triangular curve with vertices at $(0, 0)$, $(3, 0)$, and $(5, 4)$, we proceed by evaluating the integral along each segment of the triangle.

Step 1: Parameterize the three sides of the triangle

The triangle is traversed counterclockwise, starting at $(0, 0)$.

1. Segment 1: From $(0, 0)$ to $(3, 0)$

   • Equation: $y = 0$, $x \in [0, 3]$
   • $dy = 0$, so this segment contributes $0$ to the integral.

2. Segment 2: From $(3, 0)$ to $(5, 4)$

   • Parameterize as $x = 3 + 2t$, $y = 4t$, where $t \in [0, 1]$.
   • Then $dx = 2\,dt$ and $dy = 4\,dt$.
   • The integral becomes: $\int_C 3x \, dy = \int_0^1 3(3 + 2t)(4) \, dt = \int_0^1 12(3 + 2t) \, dt = \int_0^1 (36 + 24t) \, dt$ Compute: $\int_0^1 (36 + 24t) \, dt = \left[36t + 12t^2\right]_0^1 = 36(1) + 12(1)^2 - 0 = 48.$

3. Segment 3: From $(5, 4)$ to $(0, 0)$

   • Parameterize as $x = 5 - 5t$, $y = 4 - 4t$, where $t \in [0, 1]$.
   • Then $dx = -5\,dt$ and $dy = -4\,dt$.
   • The integral becomes: $\int_C 3x \, dy = \int_0^1 3(5 - 5t)(-4) \, dt = \int_0^1 -12(5 - 5t) \, dt = \int_0^1 (-60 + 60t) \, dt$ Compute: $\int_0^1 (-60 + 60t) \, dt = \left[-60t + 30t^2\right]_0^1 = -60(1) + 30(1)^2 - 0 = -30.$

Step 2: Add contributions from all segments

• Segment 1: $0$
• Segment 2: $48$
• Segment 3: $-30$

Total: $\int_C 3x \, dy = 0 + 48 - 30 = 18.$

Final Answer:

The value of the integral is $\boxed{18}$.

Do you have any questions or need more details?

Here are 5 related questions to explore:

1. How do you parameterize curves in general for line integrals?
2. Why does $dy = 0$ in the first segment of this problem?
3. What are the applications of line integrals in physics and engineering?
4. How would the integral change if the orientation of $C$ were reversed?
5. Can this integral be computed using Green's Theorem? If so, how?

Tip: For line integrals, always check if a simplification like symmetry or $dy = 0$ can make the calculation easier!