Answer
$\oint_C (x^2 y \, dx - y^2 x \, dy) = -32\pi $
Work Step by Step
Step 1: Given the line integral: \[ \oint_C (x^2 y \, dx - y^2 x \, dy) \] where \(C\) is the boundary of the region in the first quadrant enclosed between the coordinate axes and the circle \(x^2 + y^2 = 16\).
Step 2: Let \(\Phi(x, y) = x^2\) and \(\Psi(x, y) = -y^2\), and \(f(x, y) = x^2 y\) and \(g(x, y) = -y^2 x\). Using Green's Theorem: \[ \oint_C (\Phi \, dx + \Psi \, dy) = \iint_R \left(\frac{\partial \Psi}{\partial x} - \frac{\partial \Phi}{\partial y}\right) \, dA \] \[ = \iint_R (-x^2 - y^2) \, dA \] Now, let \(x = r \cos \theta\), \(y = r \sin \theta\), \(dx \, dy = r \, dr \, d\theta\). \[ \oint_C (\Phi \, dx + \Psi \, dy) = \int_0^{\pi/2} \int_0^4 (-r^3 \sin^2 \theta - r^3 \cos^2 \theta) \, r \, dr \, d\theta \] Solving this integral: \[ = -32\pi \] Result: \[ \oint_C (x^2 y \, dx - y^2 x \, dy) = -32\pi \]
