Answer
$A = \frac{ab}{2}$
Work Step by Step
The surface is bounded by the sides of the triangle with vertices: \[ (0, 0), \quad (a, 0), \quad (0, b) \quad \text{where } a, b > 0 \] The line segment from \((0, 0)\) to \((a, 0)\) can be parametrized as: \[ C_1: \quad x = t, \quad y = 0, \quad 0 \leq t \leq a \] The line segment from \((a, 0)\) to \((0, b)\) can be parametrized as: \[ C_2: \quad x = -t, \quad y = \frac{a}{b}t + b, \quad -a \leq t \leq 0 \] The line segment from \((0, b)\) to \((0, 0)\) can be parametrized as: \[ C_3: \quad x = 0, \quad y = -t, \quad -b \leq t \leq 0 \] Now, using formula (6), we have: \[ A = \oint_C x \, dy = \int_{C_1} x \, dy + \int_{C_2} x \, dy + \int_{C_3} x \, dy \] \[ = \int_0^a t \, 0 \, dt + \int_{-a}^0 (-t) \left(\frac{a}{b}t + b\right) \, dt + \int_{-b}^0 0 \, dt \] \[ = \int_{-a}^0 \left(-t^2 - \frac{ab}{b}t^2 - bt\right) \, dt \] \[ = -\frac{a}{b}\left[\frac{t^3}{3}\right]_{-a}^0 - \left[\frac{t^3}{3}\right]_{-a}^0 - \frac{b}{2}\left[t^2\right]_{-b}^0 \] \[ = -\frac{a}{b}\left(\frac{a^3}{3}\right) + \frac{a^3}{3} - \frac{b}{2}\left(\frac{b^2}{2}\right) \] \[ = -\frac{a^4}{3b} + \frac{a^3}{3} - \frac{b^3}{4} \] Finally, \[ A = \frac{ab}{2} \]
