Answer
$\Phi = 0$
Work Step by Step
The flux of the vector field \(\mathbf{F} = \mathbf{F}(x, y, z) = x\mathbf{k}\) across \(\sigma\), the portion of the paraboloid \(z = x^2 + y^2\) below the plane \(z = y\) oriented by downward unit normals is given by \[ \Phi = -\iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \,d\sigma = \iint_{R} \mathbf{F} \cdot \nabla G \,dA \] where \[ \mathbf{n} = \mathbf{n}(x, y, z) = \frac{-2x\mathbf{i} - 2y\mathbf{j} + \mathbf{k}}{\sqrt{1 + 4x^2 + 4y^2}} \] and \[ G = G(x, y, z) = z - x^2 - y^2 \] Thus, \[ \mathbf{F} \cdot \nabla G = x\mathbf{k} \cdot \left(-2x\mathbf{i} - 2y\mathbf{j} + \mathbf{k}\right) = x \] Therefore, the result is \[ \Phi = \iint_{R} x \,dA \] where \(R\) is defined as \[ R = \{(x, y) \mid x^2 + (y - 1/2)^2 = 1/4\} \] Using the parametrization \[ x = r\cos\theta \quad \text{and} \quad y = 1/2 + r\sin\theta \] we have \[ \Phi = \int_0^{\pi} \int_0^{1/2} r^2 \cos\theta \,dr\,d\theta = \int_0^{\pi} \cos\theta \,d\theta \int_0^{1/2} r^2 \,dr = 0 \] Finally, the flux of the given vector field across the surface is \[ \Phi = 0 \]
