Question

The general proof of the divergence theorem $$\int_{S} \mathbf{n} \cdot \mathbf{v} d A=\int_{V} \nabla \cdot \mathbf{v} d V$$ is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes \(x=X\) and \(X+A, y=Y\) and \(Y+B,\) and \(z=Z\) and \(Z+C,\) with total volume \(V=A B C .\) The surface \(S\) of this region is made up of six rectangles that we can call \(S_{1}\) (in the plane \(x=X\) ), \(S_{2}\) (in the plane \(x=X+A\) ), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles \(S_{1}, S_{2},\) and so forth. (a) Consider the first two of these integrals and show that $$ \int_{S_{1}} \mathbf{n} \cdot \mathbf{v} d A+\int_{S_{2}} \mathbf{n} \cdot \mathbf{v} d A=\int_{Y}^{Y+B} d y \int_{Z}^{Z+C} d z\left[v_{x}(X+A, y, z)-v_{x}(X, y, z)\right] $$ (b) Show that the integrand on the right can be rewritten as an integral of \(\partial v_{x} / \partial x\) over \(x\) running from \(x=X\) to \(x=X+A .(\text { c) Substitute the result of part }(b)\) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63).

Short answer

The process shows the equality of surface and volume integrals, proving the divergence theorem for a rectangular region.

Step-by-step solution

Consider Surface Integrals over S1 and S2

We need to evaluate the surface integrals over the two planes: - Plane 1 ( S_1 ) at x = X where \(\mathbf{n} = -\hat{i}\) because it is outward, normal to the surface.- Plane 2 ( S_2 ) at x = X+A where \(\mathbf{n} = \hat{i}\).Thus, the integrals can be written as:\[\int_{S_{1}} \mathbf{n} \cdot \mathbf{v} \, dA = -\int_{S_{1}} v_{x}(X, y, z) \, dydz \\int_{S_{2}} \mathbf{n} \cdot \mathbf{v} \, dA = \int_{S_{2}} v_{x}(X+A, y, z) \, dydz\]Adding these gives:\[\int_{S_{1}} \mathbf{n} \cdot \mathbf{v} \, dA + \int_{S_{2}} \mathbf{n} \cdot \mathbf{v} \, dA = \int_{Y}^{Y+B} dy \int_{Z}^{Z+C} dz \left[ v_{x}(X+A, y, z) - v_{x}(X, y, z) \right]\]

Rewrite using Fundamental Theorem of Calculus

The expression \(v_{x}(X+A, y, z) - v_{x}(X, y, z)\) can be rewritten using the Fundamental Theorem of Calculus in one dimension:\[v_{x}(X+A, y, z) - v_{x}(X, y, z) = \int_{X}^{X+A} \frac{\partial v_{x}}{\partial x} \, dx\]Hence, the expression from Step 1 becomes:\[\int_{Y}^{Y+B} dy \int_{Z}^{Z+C} dz \int_{X}^{X+A} \frac{\partial v_{x}}{\partial x} \, dx\]

Apply to Y and Z Planes

For the planes in y- and z- directions, similar substitutions can be done. For y planes:\[\int_{S_{3}} \mathbf{n} \cdot \mathbf{v} \, dA + \int_{S_{4}} \mathbf{n} \cdot \mathbf{v} \, dA = \int_{X}^{X+A} dx \int_{Z}^{Z+C} dz \int_{Y}^{Y+B} \frac{\partial v_{y}}{\partial y} \, dy\]And for z planes:\[\int_{S_{5}} \mathbf{n} \cdot \mathbf{v} \, dA + \int_{S_{6}} \mathbf{n} \cdot \mathbf{v} \, dA = \int_{X}^{X+A} dx \int_{Y}^{Y+B} dy \int_{Z}^{Z+C} \frac{\partial v_{z}}{\partial z} \, dz\]

Combine the Results to Prove Divergence Theorem

Add the three components for x-, y-, and z- directions. The total surface integral becomes:\[\int_{S} \mathbf{n} \cdot \mathbf{v} \, dA = \int_{X}^{X+A} dx \int_{Y}^{Y+B} dy \int_{Z}^{Z+C} \left( \frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{z}}{\partial z} \right) dz\]This is equivalent to the volume integral of the divergence:\[\int_{V} abla \cdot \mathbf{v} \, dV = \int_{V} \left( \frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{z}}{\partial z} \right) \, dV\]Therefore, the divergence theorem is confirmed.

Key concepts

Surface Integrals

Surface integrals are a way to extend the idea of integrating a function over a line to integrating over a surface. When we work with surface integrals, we are often dealing with vector fields, which are functions that assign a vector to every point in space. In the context of the Divergence Theorem, surface integrals help us measure the "flow" of a vector field across a surface.

Imagine a surface like a closed box. Each side of the box is like a window through which something might flow in or out. If we think of this flow as water or air through each side, the surface integral essentially gives us the net flow through each "window".

In the exercise, the surface is a rectangular region bounded by specific planes. We calculate the surface integrals over pairs of planes, such as those at constant values of x, y, and z.

• For each pair of opposing surfaces (like those on opposite sides of the box), surface integrals can be simplified by subtracting the values of the vector field at the two surfaces.
• This subtraction reflects the idea of net flow leaving or entering the volume enclosed by the surface.

If the vector field represents something physical, such as a fluid velocity, the surface integral tells us about the net amount of that fluid crossing the surface.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a crucial tool that connects the concept of a derivative with that of an integral. In simpler terms, it tells us that differentiation and integration are inverse processes. In one dimension, it allows us to evaluate a definite integral of a derivative function by looking at the values of the original function at the boundary of the integration range.

In the context of the given exercise, the FTC is used in a brilliant way to make the computation more manageable. When we encounter \( v_{x}(X+A, y, z) - v_{x}(X, y, z) \)in the exercise, the FTC implies we can express this difference as an integral of the derivative:\[\int_{X}^{X+A} \frac{\partial v_{x}}{\partial x} \, dx\]This expression essentially means that the change in the vector field component \(v_{x}\) across the span from \(x=X\) to \(x=X+A\) can be captured by an integral of its rate of change.

• This insight allows us to transform surface integrals into more tractable triple integrals reflecting volume.
• The power of the FTC is its ability to reduce what seems like a complex equation into something more intuitive and computation-friendly.

Using the FTC simplifies surface integrals and is integral to converting them into volume integrals, as we will see in the Divergence Theorem.

Volume Integrals

Volume integrals extend the notion of integrals into three dimensions. Instead of summing up values over a line (as with line integrals) or a surface (as with surface integrals), volume integrals consider accumulation over a three-dimensional space. This is particularly useful in capturing the totality of changes occurring throughout an entire volume, rather than just on its surface.

In this exercise, the transition from surface to volume integrals is what connects to the Divergence Theorem. The theorem itself states that the integral of a vector field's divergence over a volume \(V\) is equal to the flux through the surface that bounds \(V\). The divergence of the vector field tells us how much the field is "spreading out" per unit volume, and integrating this over the entire volume gives us a complete picture of the field's behavior.

• The divergence \( abla \cdot \mathbf{v} = \frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{z}}{\partial z} \) is key here, as it represents the rate at which the "stuff" is expanding in space.
• Calculating a volume integral of the divergence directly correlates with evaluating how field lines effectively enter or exit the enclosed region.

The Divergence Theorem is elegantly proven when we recognize this volume integral as comprising the sum of these elemental changes through the whole space, corresponding perfectly with net flows across the surface.