Divergence - Multivariable Calculus

1. Let $\mathbf{F}(x,y,z) = \langle x^3 + \sin(z), y^3 + \cos(z), R(x,y,z) \rangle$. If the vector field $\mathbf{F}$ is incompressible (i.e., has zero divergence everywhere), which of the following could be the component $R(x,y,z)$?
   Answer: $-3z(x^2+y^2) + xy$ For a vector field $\mathbf{F} = \langle P, Q, R \rangle$ to be incompressible, its divergence must be zero. $\mathrm{div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 0$. Given $P = x^3 + \sin(z)$ and $Q = y^3 + \cos(z)$, we compute their partial derivatives: $\frac{\partial P}{\partial x} = 3x^2$ $\frac{\partial Q}{\partial y} = 3y^2$ The divergence equation becomes $3x^2 + 3y^2 + \frac{\partial R}{\partial z} = 0$. This implies that the partial derivative of $R$ with respect to $z$ must be $\frac{\partial R}{\partial z} = -3x^2 - 3y^2 = -3(x^2+y^2)$. To find $R(x,y,z)$, we integrate this expression with respect to $z$: $R(x,y,z) = \int -3(x^2+y^2) \, dz = -3(x^2+y^2)z + g(x,y)$, where $g(x,y)$ is an arbitrary function of $x$ and $y$ (the 'constant' of integration with respect to $z$). We now check the answer choices to see which one has this form. A) $R(x,y,z) = -3z(x^2+y^2) + xy$. Here, $g(x,y) = xy$. This matches the required form. B) For $R = -3(x^2+y^2)$, $\frac{\partial R}{\partial z} = 0$, which is incorrect. C) For $R = z^2 - 3(x^2+y^2)z$, $\frac{\partial R}{\partial z} = 2z - 3(x^2+y^2)$, which is incorrect. D) For $R = \cos(z) - \sin(z)$, $\frac{\partial R}{\partial z} = -\sin(z) - \cos(z)$, which is incorrect.
2. Let $f(x,y,z) = e^{x+y}$ be a scalar function and $\mathbf{F}(x,y,z) = \langle y, -x, z \rangle$ be a vector field. What is the divergence of the vector field $f\mathbf{F}$?
   Answer: $e^{x+y}(1+y-x)$ We use the product rule for the divergence of a scalar function $f$ times a vector field $\mathbf{F}$: $\mathrm{div}(f\mathbf{F}) = (\nabla f) \cdot \mathbf{F} + f \mathrm{div}(\mathbf{F})$. First, we compute the required components: 1. The gradient of $f$: $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle = \langle e^{x+y}, e^{x+y}, 0 \rangle$. 2. The divergence of $\mathbf{F}$: $\mathrm{div}(\mathbf{F}) = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(-x) + \frac{\partial}{\partial z}(z) = 0 + 0 + 1 = 1$. Now, we apply the product rule formula: $(\nabla f) \cdot \mathbf{F} = \langle e^{x+y}, e^{x+y}, 0 \rangle \cdot \langle y, -x, z \rangle = y e^{x+y} - x e^{x+y} = (y-x)e^{x+y}$. $f \mathrm{div}(\mathbf{F}) = (e^{x+y})(1) = e^{x+y}$. Adding these two parts gives: $\mathrm{div}(f\mathbf{F}) = (y-x)e^{x+y} + e^{x+y} = (y-x+1)e^{x+y}$. Distractor B represents only the $(\nabla f) \cdot \mathbf{F}$ term. Distractor C represents only the $f \mathrm{div}(\mathbf{F})$ term. Distractor D might result from calculating $\mathrm{div}(\mathbf{F})=0$ incorrectly.
3. Consider a point $P=(a,b,c)$ where $a^2+b^2 > 0$. Which statement best describes the divergence of the heat flux field, $\mathrm{div}(\mathbf{H})$, at point $P$?
   Answer: $\mathrm{div}(\mathbf{H}) = 0$ because there are no heat sources or sinks at point $P$. The divergence of a vector field measures the net outflow or 'source strength' at a point. In this physical scenario, the heat is generated only along the $z$-axis (the wire). The point $P$ is specified to be not on the $z$-axis (since $a^2+b^2 > 0$). In a steady state, for any small volume around a point $P$ that does not contain a source or sink of heat, the amount of heat flowing into the volume must exactly equal the amount of heat flowing out. This is a statement of conservation of energy. Therefore, the net outflow is zero, which means the divergence of the heat flux field $\mathbf{H}$ is zero at any point $P$ not on the source line. The other options are incorrect: A is wrong because while heat flows away from the $z$-axis, $P$ itself is not a source. B is physically incorrect. D describes the magnitude of the flux field itself ($|\mathbf{H}|$), not its divergence.
4. Consider the vector field $\mathbf{G}(x,y,z) = (xyz, x^2z + y, xz^2 + y^2)$. If $S$ is the sphere $x^2 + y^2 + z^2 = 4$ and we want to find where $\text{div}(\mathbf{G})$ achieves its maximum value on $S$, which condition must be satisfied at such a point?
   Answer: $\nabla(\text{div}(\mathbf{G})) = \lambda \nabla(x^2 + y^2 + z^2 - 4)$ for some scalar $\lambda$ To find the maximum of $\text{div}(\mathbf{G})$ on the constraint surface $S$, we use Lagrange multipliers. First, $\text{div}(\mathbf{G}) = yz + 1 + x^2$. To maximize this subject to $x^2 + y^2 + z^2 = 4$, we need $\nabla(yz + 1 + x^2) = \lambda \nabla(x^2 + y^2 + z^2 - 4)$, which is equivalent to $\nabla(\text{div}(\mathbf{G})) = \lambda \nabla(x^2 + y^2 + z^2 - 4)$. Choice A incorrectly applies Lagrange multipliers to the vector field components rather than to the divergence. Choice B gives constraint equations but not optimization conditions. Choice D incorrectly relates divergence to curl and normal vector.
5. A fluid velocity field is given by $\mathbf{v}(x,y,z) = (x^2 - y^2, 2xy + z, x - z^2)$. If the divergence represents the rate of volume expansion per unit volume, at which point does the fluid exhibit the greatest compression (most negative divergence)?
   Answer: $(0, 0, 1)$ First, compute the divergence: $\text{div}(\mathbf{v}) = \frac{\partial}{\partial x}(x^2 - y^2) + \frac{\partial}{\partial y}(2xy + z) + \frac{\partial}{\partial z}(x - z^2) = 2x + 2x + (-2z) = 4x - 2z$. Evaluating at each point: At $(0,0,1)$: $4(0) - 2(1) = -2$. At $(1,0,1)$: $4(1) - 2(1) = 2$. At $(0,1,1)$: $4(0) - 2(1) = -2$. At $(1,1,0)$: $4(1) - 2(0) = 4$. The most negative divergence is $-2$, which occurs at both $(0,0,1)$ and $(0,1,1)$. Since only $(0,0,1)$ is listed as an option, it's the answer.
6. Let $\mathbf{F}(x,y,z) = \langle x^2y, y^2z, -xyz \rangle$ be a vector field. The set of all points $(x,y,z)$ where the field is incompressible (i.e., divergence-free) consists of:
   Answer: the $xz$-plane and the plane defined by $x+2z=0$. The divergence of a vector field $\mathbf{F} = \langle P, Q, R \rangle$ is given by $\mathrm{div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$. For the given field $\mathbf{F}(x,y,z) = \langle x^2y, y^2z, -xyz \rangle$, we have $P=x^2y$, $Q=y^2z$, and $R=-xyz$. The partial derivatives are $\frac{\partial P}{\partial x} = 2xy$, $\frac{\partial Q}{\partial y} = 2yz$, and $\frac{\partial R}{\partial z} = -xy$. The divergence is $\mathrm{div}(\mathbf{F}) = 2xy + 2yz - xy = xy + 2yz$. The field is incompressible where its divergence is zero. So we set $\mathrm{div}(\mathbf{F}) = 0$: $xy + 2yz = 0$. Factoring out $y$, we get $y(x+2z)=0$. This equation holds if $y=0$ or if $x+2z=0$. The equation $y=0$ describes the $xz$-plane. The equation $x+2z=0$ describes another plane. The set of points is the union of these two planes.
7. Let $\mathbf{r} = \langle x, y, z \rangle$. The gravitational field of a point mass at the origin is proportional to the vector field $\mathbf{F} = -\frac{\mathbf{r}}{\|\mathbf{r}\|^3}$. Let $\mathbf{G} = \langle x^2, y^2, z^2 \rangle$. What is the divergence of the field $\mathbf{H} = \mathbf{F} + \mathbf{G}$ at the point $(1, 2, -1)$?
   Answer: 4 We use the linearity property of the divergence operator: $\mathrm{div}(\mathbf{H}) = \mathrm{div}(\mathbf{F} + \mathbf{G}) = \mathrm{div}(\mathbf{F}) + \mathrm{div}(\mathbf{G})$. First, consider the field $\mathbf{F} = -\frac{\mathbf{r}}{\|\mathbf{r}\|^3}$. This is a standard field in physics, representing the inverse-square law for gravity or electrostatics. A key property of this field is that its divergence is zero everywhere except at the origin. So, for any point $(x,y,z) \neq (0,0,0)$, $\mathrm{div}(\mathbf{F}) = 0$. Since we are evaluating at $(1, 2, -1)$, which is not the origin, $\mathrm{div}(\mathbf{F})=0$. Next, we compute the divergence of $\mathbf{G} = \langle x^2, y^2, z^2 \rangle$. $\mathrm{div}(\mathbf{G}) = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z$. Now, we can find the divergence of $\mathbf{H}$: $\mathrm{div}(\mathbf{H}) = 0 + (2x + 2y + 2z) = 2x + 2y + 2z$. Finally, we evaluate this expression at the point $(1, 2, -1)$: $\mathrm{div}(\mathbf{H})(1,2,-1) = 2(1) + 2(2) + 2(-1) = 2 + 4 - 2 = 4$.
8. At the point $P=(3, 1, 1)$, which of the following best describes the fluid flow?
   Answer: The point is a source, with a fluid expansion rate of 12. First, we compute the divergence of the velocity field $\mathbf{v}$. Let $P=x^3, Q=y^3, R=-6xyz$. Then $\mathrm{div}(\mathbf{v}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 3x^2 + 3y^2 - 6xy$. This can be factored as $3(x^2 - 2xy + y^2) = 3(x-y)^2$. Next, we evaluate the divergence at the point $P=(3, 1, 1)$: $\mathrm{div}(\mathbf{v})(3,1,1) = 3(3-1)^2 = 3(2^2) = 12$. Since the divergence is positive (12 > 0), the point is a source, meaning there is a net outflow of fluid. The value of the divergence, 12, is the rate of expansion. Thus, the point is a source with a fluid expansion rate of 12.
9. A physical system is described by a scalar potential $\phi(x,y,z) = \ln(x^2+y^2+z^2)$. An associated vector field $\mathbf{E}$ is given by $\mathbf{E} = -\nabla \phi$. Compute the divergence of $\mathbf{E}$.
   Answer: $-\frac{2}{x^2+y^2+z^2}$ We need to compute $\mathrm{div}(\mathbf{E}) = \mathrm{div}(-\nabla \phi) = -\mathrm{div}(\nabla \phi)$. The expression $\mathrm{div}(\nabla \phi)$ is the Laplacian of $\phi$, denoted $\nabla^2 \phi$. Let $r^2 = x^2+y^2+z^2$. So $\phi = \ln(r^2) = 2\ln(r)$. First, we find the gradient of $\phi$: $\nabla \phi = \langle \frac{2x}{x^2+y^2+z^2}, \frac{2y}{x^2+y^2+z^2}, \frac{2z}{x^2+y^2+z^2} \rangle$. So, $\mathbf{E} = \langle -\frac{2x}{r^2}, -\frac{2y}{r^2}, -\frac{2z}{r^2} \rangle$. Now, we compute the divergence of $\mathbf{E}$. We need to find $\frac{\partial}{\partial x}(-\frac{2x}{r^2}) + \frac{\partial}{\partial y}(-\frac{2y}{r^2}) + \frac{\partial}{\partial z}(-\frac{2z}{r^2})$. Let's compute the first term using the quotient rule: $\frac{\partial}{\partial x}(-\frac{2x}{x^2+y^2+z^2}) = -\frac{2(x^2+y^2+z^2) - 2x(2x)}{(x^2+y^2+z^2)^2} = -\frac{2x^2+2y^2+2z^2-4x^2}{(r^2)^2} = -\frac{-2x^2+2y^2+2z^2}{r^4}$. By symmetry, the other two terms are $\frac{\partial E_y}{\partial y} = -\frac{2x^2-2y^2+2z^2}{r^4}$ and $\frac{\partial E_z}{\partial z} = -\frac{2x^2+2y^2-2z^2}{r^4}$. Summing these three terms: $\mathrm{div}(\mathbf{E}) = -\frac{1}{r^4}[(-2x^2+2y^2+2z^2) + (2x^2-2y^2+2z^2) + (2x^2+2y^2-2z^2)] = -\frac{1}{r^4}[2x^2+2y^2+2z^2] = -\frac{2r^2}{r^4} = -\frac{2}{r^2} = -\frac{2}{x^2+y^2+z^2}$.
10. For the two-dimensional vector field $\mathbf{F}(x,y) = \langle xy^2, -x^2y \rangle$, identify the region in the $xy$-plane where the flow is expanding.
    Answer: The region where $|y| > |x|$. A flow is expanding where its divergence is positive. For a 2D vector field $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$, the divergence is $\mathrm{div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$. Here, $P(x,y) = xy^2$ and $Q(x,y) = -x^2y$. Their partial derivatives are $\frac{\partial P}{\partial x} = y^2$ and $\frac{\partial Q}{\partial y} = -x^2$. So, $\mathrm{div}(\mathbf{F}) = y^2 - x^2$. The flow is expanding where $\mathrm{div}(\mathbf{F}) > 0$. This gives the inequality $y^2 - x^2 > 0$, which simplifies to $y^2 > x^2$. Taking the square root of both sides gives $\sqrt{y^2} > \sqrt{x^2}$, which is equivalent to $|y| > |x|$. This inequality describes the regions between the lines $y=x$ and $y=-x$ that contain the $y$-axis.
11. The vector field $\mathbf{H}(x,y,z) = (x\ln(y^2 + 1), ye^{xz}, z\sin(xy))$ represents a three-dimensional flow. At the point $(0, 1, \pi)$, what is the instantaneous rate of volume change per unit volume?
    Answer: $\ln(2) + e^{0 \cdot \pi} + \sin(0 \cdot 1)$ The instantaneous rate of volume change per unit volume is the divergence. Computing: $\text{div}(\mathbf{H}) = \frac{\partial}{\partial x}(x\ln(y^2 + 1)) + \frac{\partial}{\partial y}(ye^{xz}) + \frac{\partial}{\partial z}(z\sin(xy))$. The partial derivatives are: $\frac{\partial}{\partial x}(x\ln(y^2 + 1)) = \ln(y^2 + 1)$, $\frac{\partial}{\partial y}(ye^{xz}) = e^{xz}$, and $\frac{\partial}{\partial z}(z\sin(xy)) = \sin(xy)$. At $(0,1,\pi)$: $\ln(1^2 + 1) + e^{0 \cdot \pi} + \sin(0 \cdot 1) = \ln(2) + e^0 + \sin(0) = \ln(2) + 1 + 0$. Choice D correctly shows this as $\ln(2) + e^{0 \cdot \pi} + \sin(0 \cdot 1)$. The other choices have computational errors in evaluating the partial derivatives at the given point.
12. Let $\mathbf{F}(x,y,z) = (f(x,y,z), g(x,y,z), h(x,y,z))$ where $f$, $g$, and $h$ are twice continuously differentiable. If $\text{div}(\mathbf{F}) = x^2 + y^2 + z^2$ and $\text{div}(\text{curl}(\mathbf{F})) = 6$, what can be concluded?
    Answer: The given information is inconsistent because $\text{div}(\text{curl}(\mathbf{F})) \neq 0$ A fundamental theorem in vector calculus states that the divergence of any curl is always zero: $\text{div}(\text{curl}(\mathbf{F})) = 0$ for any twice continuously differentiable vector field $\mathbf{F}$. This follows from the fact that $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ due to the symmetry of mixed partial derivatives. Since we're told $\text{div}(\text{curl}(\mathbf{F})) = 6 \neq 0$, this contradicts this fundamental theorem, making the given information mathematically inconsistent. The value of $\text{div}(\mathbf{F}) = x^2 + y^2 + z^2$ is perfectly achievable (for instance, by $\mathbf{F} = (x^3/3, y^3/3, z^3/3) + \text{curl-free terms}$), so the inconsistency lies solely in the non-zero divergence of curl.
13. Let $\mathbf{F}$ be a smooth vector field on $\mathbb{R}^3$. If $\mathrm{div}(\mathbf{F})=0$ everywhere, which of the following statements must be true?
    Answer: The vector field $x\mathbf{F}$ is not necessarily divergence-free. Let's analyze each statement. A) False. $\mathrm{div}(\mathbf{F})=0$ does not imply $\mathbf{F}$ is constant. For example, $\mathbf{F} = \langle y, -x, 0 \rangle$ has $\mathrm{div}(\mathbf{F}) = 0+0+0=0$, but it is not constant. B) False. Divergence and curl are independent conditions. The same example $\mathbf{F} = \langle y, -x, 0 \rangle$ has $\mathrm{div}(\mathbf{F})=0$ but $\mathrm{curl}(\mathbf{F}) = \langle 0,0,-2 \rangle \neq \mathbf{0}$. C) False. A vector field that is the gradient of a scalar function is called conservative and satisfies $\mathrm{curl}(\mathbf{F})=0$, not $\mathrm{div}(\mathbf{F})=0$. D) True. Using the product rule: $\mathrm{div}(x\mathbf{F}) = (\nabla x) \cdot \mathbf{F} + x \mathrm{div}(\mathbf{F}) = \langle 1,0,0 \rangle \cdot \mathbf{F} + x(0) = F_1$. For $x\mathbf{F}$ to be divergence-free, we would need $F_1=0$ everywhere, which is not required by $\mathrm{div}(\mathbf{F})=0$.
14. Consider the vector field $\mathbf{G}(x,y,z) = \langle xe^{xy} - xe^{xz}, ye^{yz} - ye^{xy}, ze^{xz} - ze^{yz} \rangle$. What is the divergence of $\mathbf{G}$?
    Answer: 0 This question can be solved by recognizing a key vector identity or by direct computation. The identity is that for any twice continuously differentiable vector field $\mathbf{F}$, the divergence of its curl is always zero: $\mathrm{div}(\mathrm{curl}(\mathbf{F})) = 0$. Let's see if $\mathbf{G}$ can be expressed as the curl of some field $\mathbf{F}$. Consider $\mathbf{F} = \langle e^{yz}, e^{xz}, e^{xy} \rangle$. Its curl is $\mathrm{curl}(\mathbf{F}) = \langle \frac{\partial}{\partial y}(e^{xy}) - \frac{\partial}{\partial z}(e^{xz}), \frac{\partial}{\partial z}(e^{yz}) - \frac{\partial}{\partial x}(e^{xy}), \frac{\partial}{\partial x}(e^{xz}) - \frac{\partial}{\partial y}(e^{yz}) \rangle = \langle xe^{xy} - xe^{xz}, ye^{yz} - ye^{xy}, ze^{xz} - ze^{yz} \rangle$. This is exactly the given field $\mathbf{G}$. Since $\mathbf{G} = \mathrm{curl}(\mathbf{F})$, its divergence must be 0. Direct computation is also possible but lengthy and prone to error: $\frac{\partial}{\partial x}(xe^{xy} - xe^{xz}) = (e^{xy}+xye^{xy}) - (e^{xz}+xze^{xz})$, $\frac{\partial}{\partial y}(ye^{yz} - ye^{xy}) = (e^{yz}+yze^{yz}) - (e^{xy}+xye^{xy})$, $\frac{\partial}{\partial z}(ze^{xz} - ze^{yz}) = (e^{xz}+xze^{xz}) - (e^{yz}+yze^{yz})$. Summing these three expressions, all terms cancel out, yielding 0.

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