Answer
$\Phi = \frac{3}{2}\pi$
Work Step by Step
The flux of the vector field \(\mathbf{F}(x, y, z) = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) across the boundary of the solid bounded by the paraboloid \(z = 1 - x^2 - y^2\) and the xy-plane is given by \[ \Phi = \iiint_G \nabla \cdot \mathbf{F} \, dV \] where \[ G: \left\{ (x, y, z) \, \mid \, -1 \leq x \leq 1, -1 - x^2 \leq y \leq 1 - x^2, 0 \leq z \leq 1 - x^2 - y^2 \right\} \] Using cylindrical coordinates, this result can be expressed as \[ G: \left\{ (r, \theta, z) \, \mid \, 0 \leq \theta \leq 2\pi, 0 \leq r \leq 1, 0 \leq z \leq 1 - r^2 \right\} \] Since \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = 3 \] the flux of the vector field results in \[ \Phi = \iiint_G 3 \, dV = 3\int_0^{2\pi} \int_0^1 \int_0^{1 - r^2} r \, dz \, dr \, d\theta = 3(2\pi - 0) \int_0^1 (r - r^3) \, dr = 6\pi \left(\frac{1}{2} - \frac{1}{4}\right) = \frac{3}{2}\pi \] Finally, \[ \text{flux} = \Phi = \frac{3}{2}\pi \]
