Lesson 37: Divergence Theorem - II
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- Final exam – Tuesday, May \( 5^{\text{th}} \), 1pm – 3pm
- Cumulative (Lesson 1 – 37)
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Lesson 37: Divergence Theorem - II
Accessible transcription generated on 4/22/2026
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Review: Divergence theorem
Visual Description:
A hand-drawn technical diagram illustrating the geometric setup for the Divergence Theorem. It shows an irregular, three-dimensional solid region shaded in green, representing the volume D. This volume is completely enclosed by a blue boundary line representing the closed surface S. Several black arrows originate from different points on the blue surface and point directly away from the green interior, representing the outward-pointing normal vectors. The diagram conveys the relationship between a solid region and its oriented boundary surface.
\( S \) is a closed surface with outward normal and \( D \) is the solid region enclosed.
\[ \iiint_D \text{div} \vec{F} dV = \text{flux through boundary} = \iint_S \vec{F} \cdot d\vec{S} \]In this equation, the term \(\iiint_D \text{div} \vec{F} dV\) is a regular triple integral. To compute the surface integral \(\iint_S \vec{F} \cdot d\vec{S}\), you must parametrize \( S \):
\[ = \iint_{\substack{u,v \\ \text{Bounds}}} \vec{F}(\vec{r}(u,v)) \cdot (\vec{r}_u \times \vec{r}_v) dA \]
Note: When calculating the cross product \( (\vec{r}_u \times \vec{r}_v) \), you must choose the one pointing out to maintain the correct orientation for the outward normal.
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Example: Application of the Divergence Theorem
eg: \( S \) is the closed surface shown below
\( \vec{F} = \langle \sin z - 3x^2, 3xy, \cos x + e^y + 4xz \rangle \rightsquigarrow \text{div} \vec{F} = -6x + 3x + 4x = x \)
evaluate \( \iint_S \vec{F} \cdot d\vec{S} \)
Visual Description:
A 3D coordinate system (x, y, z axes) illustrating a solid region D. The region is bounded above by a paraboloid with the equation z = 6 - x^2 - y^2 and below by a cone with the equation z = sqrt(x^2 + y^2). The volume between these two surfaces is shaded with green diagonal hatch lines. A red vertical line segment is drawn between the lower cone surface and the upper paraboloid surface to indicate the vertical range of z at an arbitrary point (x, y). Below the 3D solid, the projection of this region onto the xy-plane is depicted as a purple shaded disk.
Applying divergence theorem \[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_D \text{div} \vec{F} \, dV = \iiint_D x \, dV \]
\[ \sqrt{x^2 + y^2} \leq z \leq 6 - x^2 - y^2 \]
\( xy \) lie in the shadow which is a disk with boundary intersection of \( 6 - x^2 - y^2 \) & \( \sqrt{x^2 + y^2} \)
Using polar: \( x^2 + y^2 = r^2 \) \[ 6 - r^2 = \sqrt{r^2} = r \] \[ \Rightarrow r^2 + r - 6 = 0 \rightsquigarrow (r+3)(r-2) = 0 \] \[ \underline{r = 2} \]
Visual Description:
A 2D coordinate system showing the xy-plane. A circle is centered at the origin with a radius labeled r = 2. The interior of the circle is shaded with purple diagonal hash marks, representing the 'shadow' or projection of the solid region onto the xy-plane. This disk defines the domain for the variables r and theta in cylindrical coordinates.
\[ \left. \begin{aligned} 0 \leq r \leq 2 \\ 0 \leq \theta \leq 2\pi \end{aligned} \right\} r \leq z \leq 6 - r^2 \]
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Calculation of Flux Using the Divergence Theorem
\[ \iint_{S} \vec{F} \cdot d\vec{s} = \iiint_{D} \operatorname{div} \vec{F} \, dV = \iiint_{D} x \, dV \]
\[ = 0 \]
Using cylindrical
\[ = \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{6 - r^2} r \cos \theta \cdot r \, dz \, dr \, d\theta \]Observe:
\[ \int_{0}^{2\pi} \cos \theta \, d\theta = \sin \theta \bigg|_{0}^{2\pi} = 0 \]Page 5
#### Flux Calculation Example
Eg: \( S \) is a sphere of radius 2, \( \vec{F} = \langle x+xz, y+yz, -x^2-y^2-z^2 \rangle \).
Compute flux.
Visual Description:
A hand-drawn diagram illustrating the problem geometry. It shows a sphere labeled 'S' in blue. The interior volume enclosed by the sphere is labeled 'D' and is indicated by green diagonal shading lines. A dashed horizontal ellipse represents the equator of the sphere, giving it a three-dimensional appearance.
\[ \text{flux} = \iint_{S} \vec{F} \cdot d\vec{s} = \iiint_{D} \text{div} \vec{F} \, dV \]
(div theorem)
\[ \text{div} \vec{F} = (1+z) + (1+z) + (-2z) = 2 \]
\[ = \iiint_{D} 2 \, dV \] \[ = 2 \iiint_{D} 1 \, dV = 2 \times \text{Volume of } D \] \[ = 2 \times \frac{4}{3} \pi (2)^3 \] \[ = \frac{64\pi}{3} \]
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Visual Description:
A 3D coordinate diagram showing a sphere centered at the origin of x, y, and z axes. The sphere's boundary is drawn in blue and labeled 'S'. The cross-section of the sphere in the x-y plane is hatched with green diagonal lines. A red vector extends from the origin to a point 'S' on the sphere's surface. The length of this vector is labeled with the Greek letter rho (\(\rho\)), and the polar angle between the positive z-axis and the vector is labeled with the Greek letter phi (\(\phi\)).
Alternative Method: Spherical Coordinates
Alternatively, use spherical coordinates to evaluate the triple integral over the solid region \( D \).
The bounds for the sphere of radius 2 are:
\[ 0 \leq \rho \leq 2 \]
\[ 0 \leq \phi \leq \pi \]
\[ 0 \leq \theta \leq 2\pi \]
Applying the Divergence Theorem in spherical coordinates:
\[ \iiint_D \text{div} \vec{F} \, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^2 \text{div} \vec{F} \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]
Given that \( \text{div} \vec{F} = 2 \), we can evaluate the iterated integral:
\[ = 2 \underbrace{\left[ 2\pi \right]}_{\text{integrate } \theta} \cdot \underbrace{\left[ 2 \right]}_{\text{integrate } \phi} \cdot \underbrace{\left[ \frac{2^3}{3} \right]}_{\text{integrate } \rho} \]
\[ = 2 \left[ 2\pi \cdot 2 \cdot \frac{8}{3} \right] = \frac{64\pi}{3} \]
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What if we remove sphere of radius 1?
Visual Description:
A 2D cross-sectional diagram illustrating a region D defined as the volume between two concentric spheres. The outer sphere is represented by a large blue circle, while the inner sphere is represented by a smaller purple circle. The annular region between these two circles is shaded with green diagonal hatching and is labeled with the letter 'D' at the top. A dashed horizontal line passes through the centers of both circles, representing their shared equatorial plane. This visualizes a spherical shell with an inner radius of 1 and an outer radius of 2.
\[ \iiint_D \text{div} \mathbf{f} \, dV \]
Bounds of D in Spherical coordinates:
\[ \begin{cases} 1 \le \rho \le 2 \\ 0 \le \phi \le \pi \\ 0 \le \theta \le 2\pi \end{cases} \] \[ \iiint_D \text{div} \vec{F} \, dV = \int_0^{2\pi} \int_0^{\pi} \int_1^2 \text{div} \vec{F} \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]Recall!
\[ \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \]
\[ \int_0^2 = \int_0^1 + \int_1^2 \implies \int_1^2 = \int_0^2 - \int_0^1 \]
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\[ \iiint_{D} \text{div} \vec{F} \, dV = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \text{div} \vec{F} \, \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta - \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \text{div} \vec{F} \, \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]
(Bounds for outer sphere) (Bounds for inner sphere)
\[ = \iiint_{\substack{\text{Sphere of} \\ \text{Radius 2}}} \text{div} \vec{F} \, dV - \iiint_{\substack{\text{Sphere of} \\ \text{Radius 1}}} \text{div} \vec{F} \, dV \]Applying stokes?
Visual Description:
A technical diagram illustrating the Divergence Theorem applied to a hollow spherical region D. The region D is the space between two concentric spheres and is shaded with green diagonal lines. The outer surface is labeled S₁ and has a red arrow pointing radially outward, indicating an outward-pointing normal vector. The inner surface is labeled S₂ and has a red arrow pointing radially inward toward the center, indicating an inward-pointing normal vector relative to the origin, which is consistent with the orientation for the boundary of the region D. The inner sphere is drawn with dashed lines to show its spherical volume.
\[ \iiint_{D} \text{div} \vec{F} \, dV = \iint_{S_1} \vec{F} \cdot d\vec{s} - \iint_{S_2} \vec{F} \cdot d\vec{s} \]
Boundary of \(D\) is:
- \(S_1\) with normal outward
- \(S_2\) with normal inward
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Divergence Theorem for Regions with Cavities
Visual Description:
A technical diagram illustrating the Divergence Theorem for a three-dimensional region with a cavity. It depicts a large blue wavy closed loop representing the outer boundary surface S1, with a red arrow indicating an outward normal vector. Nestled inside is a smaller purple wavy closed loop representing the inner boundary surface S2, with a red arrow pointing inward toward the cavity's center. The space between S1 and S2 is marked with green diagonal hatching to designate the region D. Handwritten labels clarify: 'D is region in 3D with Boundary S1 and S2', 'S1 outer boundary with Normal outward', and 'S2 inner Boundary with Normal inward'.
\( D \) is region in 3D with Boundary \( S_1 \) and \( S_2 \)
\( S_1 \): outer boundary with Normal outward
\( S_2 \): inner Boundary with Normal inward
\[ \iiint_D \operatorname{div} \vec{F} \, dV = \iint_{S_1} \vec{F} \cdot dS - \iint_{S_2} \vec{F} \cdot dS \]
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Strategy for Evaluating Surface Integrals
Want to find \(\iint_S \vec{F} \cdot d\vec{S}\)?
Is \(S\) a closed surface?
Yes:
Apply Divergence Theorem
\[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_D \text{div} \vec{F} \, dV \]
No:
Is it a curl field? i.e., \(\vec{F} = \text{curl} \vec{G}\)
Yes:
Apply Stokes' Theorem
\[ \iint_S \text{curl} \vec{G} \cdot d\vec{S} = \oint_C \vec{G} \cdot d\vec{r} \]
Visual Description:
Hand-drawn diagram illustrating a surface S and its boundary curve C. The diagram shows a three-dimensional dome-like shape representing the surface S, with curved hash marks on its interior to indicate its volume. The boundary of this surface is a closed loop at the base, labeled C. This visual setup represents the geometric relationship needed for Stokes' Theorem, where the flux of a curl through surface S is related to the line integral around its boundary curve C.
No:
Use parametrization.