# Problem Restatement **Given:** - Vector field: \( \mathbf{F} = (x, y, z) \) - Surface \( S \): Sphere \( x^2 + y^2 + z^2 = 9 \), normal pointing outwards **Task:** Use the divergence theorem to calculate the surface integral \[ \iint_S \mathbf{F} \cdot d\mathbf{A} \] --- ## **Step 1: Apply the Divergence Theorem** The **divergence theorem** states: \[ \iint_S \mathbf{F} \cdot d\mathbf{A} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV \] where \( V \) is the volume enclosed by surface \( S \). For \( \mathbf{F} = (x, y, z) \): \[ \nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3 \] **Explanation:** We replaced the surface integral with a volume integral using the divergence theorem and computed the divergence of \( \mathbf{F} \). --- ## **Step 2: Compute the Volume Integral** Since \( \nabla \cdot \mathbf{F} = 3 \), \[ \iiint_V 3 \, dV = 3 \cdot \text{(Volume of the sphere)} \] The sphere \( x^2 + y^2 + z^2 = 9 \) has radius \( r = 3 \). Volume of a sphere: \[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3^3) = \frac{4}{3} \pi \cdot 27 = 36\pi \] Multiply by 3: \[ 3 \cdot 36\pi = 108\pi \] **Explanation:** We calculated the volume of the sphere and multiplied by the divergence to get the final value. --- ## **Final Answer** \[ \boxed{108\pi} \]