Answer
$ \oint_{3x}^{2y} (3x \, dx + 2y \, dy) = 0 $
Work Step by Step
We need to determine the integral: \[ \oint_{3x}^{2y} (3x \, dx + 2y \, dy) \] along the rectangle: \[ x = -2, \quad x = 4 \] \[ y = 1, \quad y = 2 \] We have: \[P = 3xy \implies \frac{\partial p}{\partial y} = 3x\] And: \[Q = 2xy \implies \frac{\partial p}{\partial x} = 2y\] Therefore, the integral is: \[ \begin{aligned} \oint_{3x}^{2y} (3x \, dx + 2y \, dy) &= \int_{-2}^{4} \int_{1}^{2} (2y - 3x) \, dy \, dx \\ &= \int_{-2}^{4} \left[2y^2 - \frac{3}{2}xy\right] \Big|_{1}^{2} \, dx \\ &= \int_{-2}^{4} (4 - 3x - 1 + \frac{3}{2}x) \, dx \\ &= \int_{-2}^{4} (3 - \frac{3}{2}x) \, dx \\ &= \left[3x - \frac{3}{4}x^2\right]_{-2}^{4} \\ &= \left[12 - 24 + 6 + 6\right] \\ &= 0 \end{aligned} \] Result: \[ \oint_{3x}^{2y} (3x \, dx + 2y \, dy) = 0 \]
