Divergence Theorem - Multivariable Calculus

Evaluate the flux of the vector field $\mathbf{F} = \langle x^3, y^3, z^3 \rangle$ through the surface of the unit sphere $x^2+y^2+z^2=1$ .
Answer: $12\pi/5$ When you encounter flux problems through closed surfaces, immediately think about the Divergence Theorem, which converts difficult surface integrals into often-simpler volume integrals. The theorem states that $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV$ where $E$ is the region enclosed by surface $S$ . For $\mathbf{F} = \langle x^3, y^3, z^3 \rangle$ , we need the divergence: $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(z^3) = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2)$ The flux becomes: $\iiint_E 3(x^2 + y^2 + z^2) \, dV$ Converting to spherical coordinates where $x^2 + y^2 + z^2 = \rho^2$ and $dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$ : $\int_0^{2\pi} \int_0^{\pi} \int_0^1 3\rho^2 \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = 3 \int_0^{2\pi} d\theta \int_0^{\pi} \sin\phi \, d\phi \int_0^1 \rho^4 \, d\rho$ $= 3 \cdot 2\pi \cdot 2 \cdot \frac{1}{5} = \frac{12\pi}{5}$ Choice A ( $0$ ) would result from incorrectly assuming the field has zero divergence. Choice B ( $4\pi/3$ ) comes from forgetting the factor of 3 in the divergence. Choice C ( $4\pi$ ) results from miscalculating the $\rho^4$ integral as $\rho^3$ . Remember: for flux through closed surfaces, always check if the Divergence Theorem simplifies your work—it usually does when the divergence has a nice form.
Consider the vector field $\mathbf{F}(x,y,z) = (x^3, y^3, z^3)$ and the surface $S$ consisting of the part of the sphere $x^2 + y^2 + z^2 = 1$ where $z \geq 0$ , together with the disk $x^2 + y^2 \leq 1, z = 0$ . If both surfaces are oriented so their normal vectors point away from the origin, what is $\iint_S \mathbf{F} \cdot d\mathbf{S}$ ?
Answer: $\frac{6\pi}{5}$ because we apply divergence theorem to the closed hemisphere The surface $S$ is closed (hemisphere + disk), so we can apply the divergence theorem. We have $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2)$ . The region $E$ is the solid hemisphere $x^2 + y^2 + z^2 \leq 1, z \geq 0$ . Using spherical coordinates: $\iiint_E 3\rho^2 \cdot \rho^2\sin\phi \, d\rho \, d\phi \, d\theta = 3\int_0^{2\pi}\int_0^{\pi/2}\int_0^1 \rho^4\sin\phi \, d\rho \, d\phi \, d\theta = 3 \cdot 2\pi \cdot 1 \cdot \frac{1}{5} = \frac{6\pi}{5}$ . Choice A has wrong computation. Choice B uses the wrong final calculation. Choice C incorrectly assumes cancellation.
Let $E$ be the solid region in the first octant bounded by the coordinate planes, the parabolic cylinder $z = 1 - x^2$ , and the plane $y=2$ . Calculate the outward flux of $\mathbf{F} = \langle xy, y^2, xz \rangle$ through the boundary of $E$ .
Answer: $9/2$ Using the Divergence Theorem, the flux is the volume integral of the divergence. $\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(xz) = y + 2y + x = 3y+x$ . The region $E$ is described by the inequalities $0 \le x \le 1$ (since $z=1-x^2$ must be non-negative), $0 \le y \le 2$ , and $0 \le z \le 1-x^2$ . The flux is $\iiint_E (3y+x) \, dV = \int_0^1 \int_0^2 \int_0^{1-x^2} (3y+x) \, dz \, dy \, dx$ . The innermost integral (with respect to $z$ ) is $(3y+x)(1-x^2)$ . The middle integral (with respect to $y$ ) is $\int_0^2 (3y+x)(1-x^2) \, dy = (1-x^2) [\frac{3}{2}y^2 + xy]_0^2 = (1-x^2)(6+2x)$ . The outermost integral is $\int_0^1 (1-x^2)(6+2x) \, dx = \int_0^1 (6+2x-6x^2-2x^3) \, dx = [6x+x^2-2x^3-\frac{1}{2}x^4]_0^1 = 6+1-2-1/2 = 9/2$ .
Let $E$ be the solid tetrahedron with vertices at $(0,0,0)$ , $(1,0,0)$ , $(0,2,0)$ , and $(0,0,3)$ . Evaluate the outward flux of the vector field $\mathbf{F} = \langle x^2, 2y, z \rangle$ through the boundary of $E$ .
Answer: $7/2$ Using the Divergence Theorem, the flux is $\iiint_E \text{div}(\mathbf{F}) \, dV$ . First, $\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(2y) + \frac{\partial}{\partial z}(z) = 2x + 2 + 1 = 2x+3$ . The flux integral is $\iiint_E (2x+3) \, dV = \iiint_E 2x \, dV + \iiint_E 3 \, dV$ . The volume of a tetrahedron with a vertex at the origin and other vertices on the axes at $(a,0,0), (0,b,0), (0,0,c)$ is $V = abc/6$ . Here, $V = (1)(2)(3)/6 = 1$ . So, $\iiint_E 3 \, dV = 3V = 3$ . The integral $\iiint_E 2x \, dV$ is $2V\bar{x}$ , where $\bar{x}$ is the x-coordinate of the centroid of the tetrahedron. The centroid is the average of the vertices: $\bar{x} = (0+1+0+0)/4 = 1/4$ . Thus, $\iiint_E 2x \, dV = 2(1)(1/4) = 1/2$ . The total flux is $3 + 1/2 = 7/2$ .
Let $E$ be the region bounded by $z = \sqrt{x^2 + y^2}$ and $z = 2$ , and let $\mathbf{F} = (yz, xz, xy)$ . When applying the divergence theorem to find $\iint_S \mathbf{F} \cdot d\mathbf{S}$ where $S$ is the boundary of $E$ oriented outward, what is the correct setup?
Answer: $\iiint_E 0 \, dV = 0$ since $\nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0$ First, compute $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(xz) + \frac{\partial}{\partial z}(xy) = 0 + 0 + 0 = 0$ . Since the divergence is zero everywhere, the divergence theorem gives $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E 0 \, dV = 0$ . Choice A incorrectly computes the divergence. Choice C has the wrong divergence and wrong integration limits (should be from $\sqrt{x^2+y^2}$ to $2$ , or $r$ to $2$ in cylindrical). Choice D has the correct divergence but wrong limits.
A vector field $\mathbf{F}$ satisfies $\nabla \cdot \mathbf{F} = xyz$ in a region $E$ . If $E$ is the ellipsoid $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} \leq 1$ , then by the divergence theorem, $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E xyz \, dV$ where $S$ is the boundary of $E$ . What is this triple integral?
Answer: $0$ because $xyz$ is an odd function in each variable over a symmetric region The ellipsoid $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} \leq 1$ is symmetric about each coordinate plane (the $yz$ -plane, $xz$ -plane, and $xy$ -plane). The function $f(x,y,z) = xyz$ is odd in $x$ : $f(-x,y,z) = -xyz = -f(x,y,z)$ . Similarly, it's odd in $y$ and $z$ . When integrating an odd function over a region symmetric about the origin, the integral equals zero because the contributions from opposite sides cancel. Choice B incorrectly assumes a non-zero average value. Choices C and D ignore the symmetry property.
Let $S_1$ be the lateral surface of the cone $z = \sqrt{x^2+y^2}$ for $0 \le z \le 2$ , oriented upward. Let $S_2$ be the disk $x^2+y^2 \le 4$ in the plane $z=2$ , also oriented upward. For the vector field $\mathbf{F} = \langle 3x-y, 3y+z, 3z-x \rangle$ , what is the flux $\iint_{S_1} \mathbf{F} \cdot d\mathbf{S}$ ?
Answer: $0$ Let $E$ be the solid cone bounded by $S_1$ and $S_2$ . The boundary of $E$ with outward orientation is $S_{out} = S_{2,up} \cup S_{1,down}$ . By the Divergence Theorem, the total outward flux is $\iiint_E \text{div}(\mathbf{F}) \, dV$ . Here, $\text{div}(\mathbf{F}) = 3+3+3=9$ . The volume of the cone is $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2^2)(2) = 8\pi/3$ . The total flux is $9V = 9(8\pi/3) = 24\pi$ . So, $\iint_{S_{2,up}} \mathbf{F} \cdot d\mathbf{S} + \iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 24\pi$ . Let's calculate the flux through the top disk $S_2$ . The normal is $\mathbf{n}=\langle 0,0,1 \rangle$ . On $S_2$ , $z=2$ , so $\mathbf{F} = \langle 3x-y, 3y+2, 6-x \rangle$ . The flux is $\iint_{S_2} (6-x) \, dA = 6 \cdot \text{Area}(S_2) - \iint_{S_2} x \, dA = 6(\pi 2^2) - 0 = 24\pi$ . Substituting this into the Divergence Theorem equation: $24\pi + \iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 24\pi$ , which implies $\iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 0$ . The question asks for the flux through $S_1$ oriented upward, which is the opposite orientation: $\iint_{S_{1,up}} \mathbf{F} \cdot d\mathbf{S} = -\iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = -0 = 0$ .
The divergence theorem is used to convert the surface integral $\iint_S (x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}) \cdot d\mathbf{S}$ over the closed surface $S$ bounding region $E$ into $\iiint_E 2(x+y+z) \, dV$ . If this conversion is correct and $E$ is the unit cube $[0,1]^3$ , what constraint must $S$ satisfy?
Answer: $S$ can be any closed surface bounding a region $E$ since $\nabla \cdot (x^2, y^2, z^2) = 2x + 2y + 2z = 2(x+y+z)$ everywhere Let's verify the divergence: $\nabla \cdot (x^2, y^2, z^2) = \frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z} = 2x + 2y + 2z = 2(x+y+z)$ . This computation is valid everywhere in $\mathbb{R}^3$ , so the divergence theorem applies to any closed, smooth, orientable surface $S$ bounding any region $E$ . The specific choice of $E = [0,1]^3$ is just for evaluation purposes. Choice A incorrectly restricts the domain. Choice C incorrectly suggests the divergence depends on the origin. Choice D is partially correct but misses that the divergence is $2(x+y+z)$ everywhere, not just in special regions.
Let $\mathbf{F} = \langle e^{y^2}\sin(z) + 3x, x^5\cos(z) - 2y, \ln(x^2+y^2+1) + 5z \rangle$ . What is the outward flux of $\mathbf{F}$ through the surface of the cube with vertices at $(\pm 1, \pm 1, \pm 1)$ ?
Answer: $48$ Directly calculating the flux over the six faces of the cube would be extremely complicated. The Divergence Theorem is the appropriate tool. We first compute the divergence of $\mathbf{F}$ : $\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(e^{y^2}\sin(z) + 3x) + \frac{\partial}{\partial y}(x^5\cos(z) - 2y) + \frac{\partial}{\partial z}(\ln(x^2+y^2+1) + 5z) = 3 - 2 + 5 = 6$ . Since the divergence is a constant, the flux is $\iiint_E 6 \, dV = 6 \cdot \text{Volume}(E)$ . The cube has vertices at $(\pm 1, \pm 1, \pm 1)$ , so its side length is 2. The volume is $2^3 = 8$ . Therefore, the flux is $6 \times 8 = 48$ .
Let $E$ be the region between the concentric spheres $x^2+y^2+z^2=1$ and $x^2+y^2+z^2=4$ . Find the total outward flux of the vector field $\mathbf{F} = \frac{\mathbf{r}}{|\mathbf{r}|^3}$ through the boundary of $E$ , where $\mathbf{r} = \langle x,y,z \rangle$ .
Answer: $0$ The vector field is $\mathbf{F} = \frac{\langle x, y, z \rangle}{(x^2+y^2+z^2)^{3/2}}$ . The divergence of this field is $\text{div}(\mathbf{F}) = 0$ for all $\mathbf{r} \neq \mathbf{0}$ . The region $E$ is defined by $1 \le x^2+y^2+z^2 \le 4$ , which does not contain the origin. Since $\mathbf{F}$ is continuously differentiable on $E$ and its boundary, we can apply the Divergence Theorem. The total outward flux is $\iiint_E \text{div}(\mathbf{F}) \, dV = \iiint_E 0 \, dV = 0$ . Alternatively, one can calculate the flux through the outer sphere (radius 2, outward normal) and the inner sphere (radius 1, inward normal relative to origin). The flux through a sphere of radius $R$ is $4\pi$ . For the outer sphere, the flux is $4\pi$ . For the inner sphere, the outward normal from the region points towards the origin, yielding a flux of $-4\pi$ . The sum is $4\pi - 4\pi = 0$ .
Let $\mathbf{F}$ be a vector field with constant divergence $\text{div}(\mathbf{F}) = k > 0$ . Let $S_1$ be the surface of a sphere of radius $R$ and $S_2$ be the surface of a cube of side length $2R$ , both centered at the origin. Let $\Phi_1$ and $\Phi_2$ be the outward fluxes through $S_1$ and $S_2$ respectively. Which of the following statements is true?
Answer: $\Phi_1 < \Phi_2$ According to the Divergence Theorem, the flux of a vector field through a closed surface is the triple integral of its divergence over the enclosed volume. Since the divergence is a constant $k$ , the flux is $\Phi = \iiint_E k \, dV = k \cdot \text{Volume}(E)$ . For the sphere $S_1$ , the enclosed volume is $V_1 = \frac{4}{3}\pi R^3$ . So, $\Phi_1 = k \cdot \frac{4}{3}\pi R^3$ . For the cube $S_2$ with side length $2R$ , the enclosed volume is $V_2 = (2R)^3 = 8R^3$ . So, $\Phi_2 = k \cdot 8R^3$ . To compare $\Phi_1$ and $\Phi_2$ , we compare their volumes. We need to compare $\frac{4}{3}\pi$ with 8. Since $\pi \approx 3.14159$ , $\frac{4}{3}\pi \approx 4.189$ . As $4.189 < 8$ , we have $V_1 < V_2$ . Since $k>0$ , it follows that $\Phi_1 < \Phi_2$ .
Let $E$ be the region defined by $1 \le x^2+y^2+z^2 \le 4$ . The outward flux of a radial vector field $\mathbf{F} = f(\rho)\mathbf{e}_\rho$ through the boundary of $E$ is $14\pi$ . Here, $\rho = \sqrt{x^2+y^2+z^2}$ and $\mathbf{e}_\rho$ is the radial unit vector. Which of the following could be the function $f(\rho)$ ?
Answer: $f(\rho) = \rho/2$ The flux can be found using the Divergence Theorem or by direct integration over the two spherical surfaces. Using direct integration, the flux through a sphere of radius $\rho$ is $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S f(\rho)\mathbf{e}_\rho \cdot \mathbf{e}_\rho \, dA = f(\rho) \cdot (\text{Surface Area}) = f(\rho) \cdot 4\pi\rho^2$ . The total outward flux from the region $E$ is the flux through the outer sphere (radius 2) minus the flux through the inner sphere (radius 1). Flux = $f(2) \cdot 4\pi(2^2) - f(1) \cdot 4\pi(1^2) = 16\pi f(2) - 4\pi f(1) = 4\pi(4f(2) - f(1))$ . We are given that the flux is $14\pi$ . So, $4\pi(4f(2) - f(1)) = 14\pi$ , which simplifies to $4f(2) - f(1) = 14/4 = 7/2$ . We test the given options: For $f(\rho)=\rho/2$ , we have $4f(2) - f(1) = 4(2/2) - (1/2) = 4(1) - 1/2 = 7/2$ . This matches the condition.
Let $S$ be the portion of the paraboloid $z = 4 - x^2 - y^2$ for $z \ge 0$ , with upward orientation. Evaluate the flux integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$ for the vector field $\mathbf{F}(x,y,z) = \langle xz, yz, x^2 + y^2 \rangle$ .
Answer: $88\pi/3$ The surface $S$ is open. To use the Divergence Theorem, we close the surface with the disk $S_{cap}$ defined by $x^2+y^2 \le 4$ at $z=0$ , oriented downward to ensure the total orientation is outward from the enclosed solid $E$ . The total flux through the closed surface $S \cup S_{cap}$ is given by $\iiint_E \text{div}(\mathbf{F}) \, dV$ . The divergence of $\mathbf{F}$ is $\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(x^2+y^2) = z + z + 0 = 2z$ . Using cylindrical coordinates, the volume integral is $\int_0^{2\pi} \int_0^2 \int_0^{4-r^2} (2z) r \, dz \, dr \, d\theta = 2\pi \int_0^2 r[z^2]_0^{4-r^2} \, dr = 2\pi \int_0^2 r(4-r^2)^2 \, dr = 64\pi/3$ . This is the total flux. The flux through the cap $S_{cap}$ (where $z=0$ and the normal is $\mathbf{n}=\langle 0,0,-1 \rangle$ ) is $\iint_{S_{cap}} \langle 0, 0, x^2+y^2 \rangle \cdot \langle 0,0,-1 \rangle \, dA = \iint_{S_{cap}} -(x^2+y^2) \, dA = -\int_0^{2\pi}\int_0^2 r^2 \cdot r \, dr \, d\theta = -8\pi$ . The flux through $S$ is the total flux minus the flux through the cap: $64\pi/3 - (-8\pi) = 88\pi/3$ .
Let $\mathbf{F}$ be a continuously differentiable vector field in $\mathbb{R}^3$ . The flux of $\mathbf{F}$ out of any sphere centered at the origin is found to be proportional to the cube of the sphere's radius $R$ , i.e., $\Phi(R) = CR^3$ for some constant $C$ . What is the divergence of $\mathbf{F}$ at the origin, $\text{div}(\mathbf{F})(0,0,0)$ ?
Answer: $3C/(4\pi)$ By the Divergence Theorem, the flux $\Phi(R)$ through a sphere of radius $R$ is equal to the triple integral of its divergence over the enclosed ball $B_R$ . So, $\iiint_{B_R} \text{div}(\mathbf{F}) \, dV = CR^3$ . By the Mean Value Theorem for integrals, there exists a point $P$ in $B_R$ such that $\text{div}(\mathbf{F})(P) \cdot \text{Volume}(B_R) = CR^3$ . The volume of the ball is $\frac{4}{3}\pi R^3$ . Thus, $\text{div}(\mathbf{F})(P) \cdot \frac{4}{3}\pi R^3 = CR^3$ , which simplifies to $\text{div}(\mathbf{F})(P) = \frac{3C}{4\pi}$ . As we take the limit $R \to 0$ , the point $P$ must approach the origin. Since $\mathbf{F}$ is continuously differentiable, its divergence is continuous, so $\text{div}(\mathbf{F})(0,0,0) = \lim_{R\to 0} \text{div}(\mathbf{F})(P) = \frac{3C}{4\pi}$ .

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