Question

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iiint \nabla \cdot \mathbf{V} d \tau\) over the unit cube in the first octant, where $$\mathbf{V}=\left(x^{3}-x^{2}\right) y \mathbf{i}+\left(y^{3}-2 y^{2}+y\right) x \mathbf{j}+\left(z^{2}-1\right) \mathbf{k}$$.

Short answer

The integral evaluates to 1.

Step-by-step solution

Identify the integral

The given integral is a volume integral: \[ \iiint abla \cdot \mathbf{V} d \tau \] over the unit cube in the first octant.

Determine the divergence

Calculate the divergence of the vector field \(\mathbf{V}\). \[ abla \cdot \mathbf{V} = \frac{\partial}{\partial x} \left( (x^3 - x^2)y \right) + \frac{\partial}{\partial y} \left( (y^3 - 2y^2 + y)x \right) + \frac{\partial}{\partial z} \left( z^2 - 1 \right) \] Compute each term separately: \[ \frac{\partial}{\partial x} \left( (x^3 - x^2)y \right) = y \cdot \frac{d}{dx} (x^3 - x^2) = y (3x^2 - 2x) \] \[ \frac{\partial}{\partial y} \left( (y^3 - 2y^2 + y)x \right) = x \cdot \frac{d}{dy} (y^3 - 2y^2 + y) = x (3y^2 - 4y + 1) \] \[ \frac{\partial}{\partial z} \left( z^2 - 1 \right) = 2z \]

Combine the results

Combine the results of the partial derivatives: \[ abla \cdot \mathbf{V} = y (3x^2 - 2x) + x (3y^2 - 4y + 1) + 2z \]

Set up the volume integral

Set up the volume integral over the unit cube \([0,1] \times [0,1] \times [0,1]\) in the first octant: \[ \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} \left[ y (3x^2 - 2x) + x (3y^2 - 4y + 1) + 2z \right] dx\, dy \ dz \]

Solve the integral

Evaluate the integral part by part: 1. \( \int_{0}^{1} \left(\int_{0}^{1} \int_{0}^{1} y (3x^2 - 2x) \, dx \, dy \, dz \right) \) 2. \( \int_{0}^{1} \left( \int_{0}^{1} \int_{0}^{1} x (3y^2 - 4y + 1) \, dx \, dy \, dz \right) \) 3. \( \int_{0}^{1} \left( \int_{0}^{1} \int_{0}^{1} 2z \, dx \, dy \, dz \right) \)

Evaluate each term

Solve each part separately: 1. \( y \left[ \int_{0}^{1} (3x^2 - 2x) \ dx = \left[ x^3 - x^2 \right]_{0}^{1} = 1 - 1 = 0 \right] \times \int_{0}^{1} y \, dy \times \int_{0}^{1} dz = 0 \) 2. \( x \left[ \int_{0}^{1} (3y^2 - 4y + 1) \ dy = \left[ y^3 - 2y^2 + y \right]_{0}^{1} = 1 - 2 + 1 = 0 \right] \times \int_{0}^{1} dx \times \int_{0}^{1} dz = 0 \) 3. \( 2z \left[ \int_{0}^{1} \int_{0}^{1} 1 \, dx \, dy = 1 \right] \times \int_{0}^{1} z \ dz = 2 \times \left[ \frac{z^2}{2} \right]_{0}^{1} = 1 \)

Combine the evaluated parts

Add the evaluated parts together: \[ 0 + 0 + 1 = 1 \]

Key concepts

divergence theorem

The divergence theorem is a critical concept in vector calculus. It connects surface integrals with volume integrals. In essence, it states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of that field within the surface. Mathematically, it is expressed as: ■ To better grasp this concept, let's look at our problem. We evaluated a volume integral, specifically: ■ This process applied the divergence theorem, confirming its utility in simplifying complex volume calculations. -> Consider practicing this with various vector fields and geometries to master its utility!

surface integrals

Surface integrals are integral calculus tools that evaluate functions over a surface. In vector calculus, they typically involve vector fields. For instance, integrating In our exercise, we could have chosen a surface instead of volume. The surface integral:

■ Surface integrals help us understand the flow of a vector field over a surface.

applied the Divergence Theorem, showing simplicity and beauty in translating higher-dimensional concepts into manageable parts. ■ One key difference from volume integrals is potential integrals change from parameter to parameter!
Seek different surfaces to apply this idea and improve your understanding.

volume integrals

Volume integrals are used to integrate over a three-dimensional region. They are particularly useful when dealing with scalar fields or vector fields defined over a volume, such as our unit cube in the first octant. The integral we studied aimed at solving:
Each integral part (dx, dy, dz) considers all volumetric units, summing up to the desired whole. This technique helps solve real-world problems such as fluid dynamics, electromagnetism, and more.
Don’t hesitate to break the process down:

■ Identify limits over x, y, z axes, & do one step at a time.

Understanding these minor steps collectively builds confidence and clarity.

partial derivatives

Partial derivatives measure how a multivariable function changes as one of its variables changes while keeping others constant. In our vector function

In the given exercise:
Partial derivatives were calculated for -> z as constant, making the process easier.
Practice finding derivatives:
Identifying these derivatives simplifies vector calculus significantly!