Questions: 10. Let F = ⟨3x² - 4y², -2xyz, 4z²⟩ be the velocity field of a fluid. (a) (2 points) Find div F. (b) (5 points) We wish to use the Divergence Theorem to evaluate ∬ₛ F ⋅ dS, where S is the surface of the portion of the paraboloid z = R² - x² - y², where R > 0, with y ≥ 0 and z ≥ 0. All three surfaces of the solid are included in S, and S is oriented outward. You need to i. sketch the region S (making sure to label your axes); and ii. set up the integral obtained by applying the Divergence Theorem to compute the integral. Make sure to obtain all necessary limits and simplify the integrand. You do NOT need to evaluate the resulting integral.

(a) Find \(\operatorname{div} \mathbf{F}\).

Calculate the divergence of \(\mathbf{F}\).

The divergence of a vector field \(\mathbf{F} = \langle P, Q, R \rangle\) is given by \(\operatorname{div} \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\).

For \(\mathbf{F} = \langle 3x^2 - 4y^2, -2xyz, 4z^2 \rangle\):

\[ \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(3x^2 - 4y^2) = 6x \]

\[ \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(-2xyz) = -2xz \]

\[ \frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(4z^2) = 8z \]

Thus, \(\operatorname{div} \mathbf{F} = 6x - 2xz + 8z\).

\(\boxed{\operatorname{div} \mathbf{F} = 6x - 2xz + 8z}\)

(b) Use the Divergence Theorem to evaluate \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\).

Sketch the region \(S\).

The region \(S\) is the surface of the paraboloid \(z = R^2 - x^2 - y^2\) with \(y \geq 0\) and \(z \geq 0\). This is a paraboloid opening downward, truncated by the plane \(z = 0\) and the plane \(y = 0\). The region is symmetric about the \(x\)-axis and lies in the first and fourth quadrants of the \(xy\)-plane.

Set up the integral using the Divergence Theorem.

The Divergence Theorem states:

\[ \iint_{S} \mathbf{F} \cdot d \mathbf{S} = \iiint_{V} \operatorname{div} \mathbf{F} \, dV \]

where \(V\) is the volume enclosed by \(S\).

The volume \(V\) is bounded by \(z = R^2 - x^2 - y^2\), \(y \geq 0\), and \(z \geq 0\).

In cylindrical coordinates, \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\), the limits are:

• \(0 \leq r \leq R\)
• \(0 \leq \theta \leq \pi\)
• \(0 \leq z \leq R^2 - r^2\)

The divergence \(\operatorname{div} \mathbf{F} = 6x - 2xz + 8z\) in cylindrical coordinates becomes:

\[ 6r\cos\theta - 2r\cos\theta z + 8z \]

The integral is:

\[ \iiint_{V} (6r\cos\theta - 2r\cos\theta z + 8z) \, r \, dz \, dr \, d\theta \]

Simplifying the integrand:

\[ = \iiint_{V} (6r^2\cos\theta - 2r^2\cos\theta z + 8rz) \, dz \, dr \, d\theta \]

This is the integral to be evaluated.

\(\boxed{\iiint_{V} (6r^2\cos\theta - 2r^2\cos\theta z + 8rz) \, dz \, dr \, d\theta}\)

\(\boxed{\operatorname{div} \mathbf{F} = 6x - 2xz + 8z}\) \(\boxed{\iiint_{V} (6r^2\cos\theta - 2r^2\cos\theta z + 8rz) \, dz \, dr \, d\theta}\)