Question

Integration by parts (Gauss' Formula) Recall the Product Rule of Theorem \(17.13: \nabla \cdot(u \mathbf{F})=\nabla u \cdot \mathbf{F}+u(\nabla \cdot \mathbf{F})\) a. Integrate both sides of this identity over a solid region \(D\) with a closed boundary \(S\), and use the Divergence Theorem to prove an integration by parts rule: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V$$ b. Explain the correspondence between this rule and the integration by parts rule for single-variable functions. c. Use integration by parts to evaluate \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\) where \(D\) is the cube in the first octant cut by the planes \(x=1\) \(y=1,\) and \(z=1\)

Short answer

#Answer# a) Proving the integration by parts rule for the given identity, we have: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V.$$ b) The correspondence between the single-variable and multivariable integration by parts rules is that they express the same concept but for different types of domains and functions. In both cases, they express the integration of the product of two functions in terms of the product evaluation at the boundary and the subtraction of an integral involving their derivatives. c) The given triple integral evaluates to: $$\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V = \frac{1}{16}.$$

Step-by-step solution

Integrate both sides of the identity

To prove the integration by parts rule, we first integrate both sides of the given identity \(\nabla \cdot(u \mathbf{F})=\nabla u \cdot \mathbf{F}+u(\nabla \cdot \mathbf{F})\) over the region \(D\), which gives us $$\iiint_{D} (\nabla \cdot(u \mathbf{F})) dV = \iiint_{D} (\nabla u \cdot \mathbf{F})dV + \iiint_{D} (u (\nabla \cdot \mathbf{F}))dV.$$ Step 2: Apply the Divergence Theorem

Apply the Divergence Theorem

Now, we apply the Divergence Theorem on the left-hand side of the equation. The Divergence Theorem states that $$\iiint_{D} (\nabla \cdot \mathbf{G}) dV = \iint_{S} \mathbf{G} \cdot \mathbf{n} dS$$ Applying the theorem to our equation with \(\mathbf{G} = u\mathbf{F}\), we get $$\iint_{S} (u \mathbf{F} \cdot \mathbf{n}) dS = \iiint_{D} (\nabla u \cdot \mathbf{F})dV + \iiint_{D} (u (\nabla \cdot \mathbf{F}))dV.$$ Step 3: Rearrange terms to get the Integration by Parts Rule

Rearrange terms to get the Integration by Parts Rule

Finally, we rearrange the terms to obtain the desired integration by parts rule: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V$$ #Part b: Correspondence with Single-Variable Rule#

Correspondence with Single-Variable Rule

In the single-variable integration by parts rule, we have $$\int_{a}^{b} u(x)v'(x)dx = u(x)v(x)\big|_{a}^{b} - \int_{a}^{b}u'(x)v(x)dx.$$ Comparing this with our multivariate rule, we can see that the term \(\iiint_{D} u(\nabla \cdot \mathbf{F}) d V\) corresponds to \(\int_{a}^{b} u(x)v'(x)dx\), and the term \(\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S\) corresponds to the evaluation of the product \(u(x)v(x)\) at the boundary. Finally, the term \(-\iiint_{D} \nabla u \cdot \mathbf{F} d V\) corresponds to the subtraction of the integral of the product of the derivatives, \(-\int_{a}^{b}u'(x)v(x)dx\). Thus, both rules express the same integration by parts concept, but for different types of domains and functions. #Part c: Evaluate the Given Triple Integral#

Identify u and F

We are given the triple integral \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\) and we need to apply the integration by parts rule. We first identify the scalar function \(u\) and the vector field \(\mathbf{F}\) such that the given integral is of the form \(\iiint_{D} u(\nabla \cdot \mathbf{F}) d V\). In this case, \(u = x^2y + y^2z + z^2x\), and we need to choose \(\mathbf{F}\) such that \(\nabla \cdot \mathbf{F} = u\). Step 1: Choose \(\mathbf{F}\)

Choose \(\mathbf{F}\)

Let's choose a simple vector field \(\mathbf{F} = (0, x^2z, y^2x)\). This vector field has a divergence of \(\nabla \cdot \mathbf{F} = x^2y + y^2z + z^2x\), which matches our given \(u\). Step 2: Apply the Integration by Parts Rule

Apply the Integration by Parts Rule

Now we apply the integration by parts rule to evaluate the given triple integral: $$\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V.$$ Step 3: Evaluate Surface Integral

Evaluate Surface Integral

The surface of the cube consists of six faces, each with outward normal vectors \(\mathbf{n} = \pm\mathbf{i}\), \(\pm\mathbf{j}\), and \(\pm\mathbf{k}\), respectively. For these faces, \(x=0\), \(x=1\), \(y=0\), \(y=1\), \(z=0\), and \(z=1\). On these faces, the integral of \(u\mathbf{F}\cdot\mathbf{n}\) becomes zero, because \(u=x^2y+y^2z+z^2x\) vanishes on the faces, and hence the surface integral becomes $$\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S=0.$$ Step 4: Evaluate Volume Integral

Evaluate Volume Integral

Next, we evaluate the volume integral. We find the gradient of \(u\): $$\nabla u = (2xy + z^2, x^2 + 2yz, 2zx + y^2).$$ Dotting this with \(\mathbf{F}\), we have $$\nabla u \cdot \mathbf{F} = 0(x^2z) + (x^2 + 2yz)(y^2x) + 0(2zx + y^2) = x^3y^3.$$ So the volume integral is: $$-\iiint_{D} \nabla u \cdot \mathbf{F} d V = -\iiint_{D} x^3y^3 dV.$$ Step 5: Evaluate the Remaining Integral

Evaluate the Remaining Integral

Now we just need to evaluate the integral over the cube \(D\): $$-\iiint_{D} x^3y^3dV = -\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} x^3y^3 dz dy dx= -\int_{0}^{1}\int_{0}^{1}(x^3y^3)\Big|_{0}^{1} dy dx,$$ which simplifies to: $$-\int_{0}^{1}\int_{0}^{1} x^3y^3 dy dx = -\int_{0}^{1}x^3\int_{0}^{1}y^3 dy dx = -\int_{0}^{1}x^3(\frac{1}{4}) dx = -\frac{1}{4}\int_{0}^{1}x^3 dx.$$ Now we evaluate the remaining integral, which leads to: $$-\frac{1}{4}\int_{0}^{1}x^3 dx = -\frac{1}{4}\left(\frac{1}{4}\right)\Big|_0^1 = -\frac{1}{16}.$$ Step 6: Plug the Results in the Integration by Parts Rule

Plug the Results in the Integration by Parts Rule

Finally, substituting our surface and volume integral results into the integration by parts rule for our given integral, we obtain $$\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V = 0 - \left(-\frac{1}{16}\right) = \boxed{\frac{1}{16}}.$$

Key concepts

Divergence Theorem

The Divergence Theorem is a crucial concept in vector calculus. It connects the flow of a vector field across a closed surface to the behavior of the vector field inside the surface's boundary.
This theorem is often expressed as:

• \[ \iiint_{D} (abla \cdot \mathbf{G}) \, dV = \iint_{S} \mathbf{G} \cdot \mathbf{n} \, dS \]

Here, \( D \) is a solid region, \( S \) is the closed surface bounding \( D \), \( \mathbf{G} \) is a vector field, and \( \mathbf{n} \) is the outward normal vector on \( S \).
Essentially, the Divergence Theorem states that the integral of the divergence of a vector field over a volume equals the integral of the field's normal component over the volume's boundary.
This powerful tool allows us to switch between volume integrals and surface integrals, making it easier to solve complex problems in physics and engineering.

Product Rule

The Product Rule in calculus is a fundamental principle. It describes how to differentiate the product of two functions. For single variable functions \( u(x) \) and \( v(x) \), it is given by:

• \( (uv)' = u'v + uv' \)

In the context of vector calculus, a similar rule applies to the dot product of a scalar function \( u \) and a vector field \( \mathbf{F} \):

• \( abla \cdot (u \mathbf{F}) = abla u \cdot \mathbf{F} + u(abla \cdot \mathbf{F}) \)

The expression \( abla u \cdot \mathbf{F} \) represents the dot product of the gradient of \( u \) with vector field \( \mathbf{F} \), while \( u(abla \cdot \mathbf{F}) \) considers \( u \) scaling the divergence of \( \mathbf{F} \).
This rule is very important in deriving integration by parts in multiple dimensions using the Divergence Theorem.

Triple Integral

Triple integrals extend the idea of integration to functions of three variables. They are used to calculate volumes and other properties of a three-dimensional region \( D \).
Mathematically, a triple integral is represented by:

• \( \iiint_{D} f(x, y, z) \, dV \)

Where \( dV \) represents a differential volume element. The triple integral sums up infinitesimal contributions across the entire region \( D \), allowing us to compute total volumes or other cumulative measures.
These integrals are vital in physics for computing mass, charge, or energy densities across a 3D space.
When using triple integrals, it's crucial to set up bounds carefully and identify the integration limits for each variable based on the region's geometric constraints.

Vector Field

A vector field assigns a vector to every point in space. It represents various physical quantities such as velocity in fluid dynamics or force fields in electromagnetics.
In mathematical terms, a vector field \( \mathbf{F} \) in three dimensions is represented as:

• \( \mathbf{F}(x, y, z) = P(x, y, z) \mathbf{i} + Q(x, y, z) \mathbf{j} + R(x, y, z) \mathbf{k} \)

Where \( P, Q, R \) define the components of the vector at each point and \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors in the x, y, and z directions.
Vector fields can be visualized as arrows varying in direction and magnitude at every point. They are especially useful in visualizing patterns of flow and are fundamental to applying theorems like the Divergence Theorem.
In applications, understanding the divergence, curl, and gradient related to vector fields helps in analyzing flow behavior and other vector-related phenomena.