Question

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the entire surface of the volume in the first octant bounded by \(x^{2}+y^{2}+z^{2}=16\) and the coordinate planes, where $$\mathbf{V}=\left(x+x^{2}-y^{2}\right) \mathbf{i}+(2 x y z-2 x y) \mathbf{j}-x z^{2} \mathbf{k}$$

Short answer

\[\frac{32\pi}{3}\]

Step-by-step solution

- Identify the Integral and Theorem to Use

The given integral is a surface integral of the vector field \(\mathbf{V}\cdot\mathbf{n}\,d\sigma\)over the surface in the first octant bounded by \(x^2 + y^2 + z^2 = 16\) and the coordinate planes. Since the integral involves a surface in three-dimensional space, the Divergence Theorem will be appropriate here.

- State the Divergence Theorem

The Divergence Theorem states: \[\iint_S \mathbf{V}\cdot\mathbf{n}\,d\sigma = \iiint_V abla\cdot \mathbf{V}\,dV\] where \(abla \cdot \mathbf{V}\) denotes the divergence of \(\mathbf{V}\), and the right side of the equation is a volume integral over the volume \(V\) enclosed by the surface \(S\).

- Calculate the Divergence \(abla\cdot \mathbf{V}\)

Given \(\mathbf{V} = (x+x^2-y^2)\mathbf{i} + (2xyz - 2xy)\mathbf{j} - xz^2\mathbf{k}\), calculate \(abla\cdot \mathbf{V}\). To find the divergence, compute: \[abla \cdot \mathbf{V} = \frac{\partial}{\partial x}(x + x^2 - y^2) + \frac{\partial}{\partial y}(2xyz - 2xy) + \frac{\partial}{\partial z}(-xz^2)\] Evaluate each term:\[\frac{\partial}{\partial x}(x + x^2 - y^2) = 1 + 2x\] \[\frac{\partial}{\partial y}(2xyz - 2xy) = 2xz - 2x\] \[\frac{\partial}{\partial z}(-xz^2) = -2xz\] Sum these results:\[abla \cdot \mathbf{V} = (1 + 2x) + (2xz - 2x) + (-2xz) = 1\]

- Set Up the Volume Integral

According to the Divergence Theorem, \[\iint_S \mathbf{V}\cdot\mathbf{n}\,d\sigma = \iiint_V 1\,dV\] Since the integrand is 1, we just need the volume of the region bounded by \(x^2 + y^2 + z^2 = 16\) in the first octant.

- Determine the Volume in the First Octant

The volume of the sphere \(x^2 + y^2 + z^2 = 16\) is \(\frac{4}{3} \pi r^3\) where the radius \(r = 4\). Total volume: \[\frac{4}{3} \pi (4)^3 = \frac{256\pi}{3}\] Since we only need the volume in the first octant, divide by 8: \[\frac{256\pi}{3 \cdot 8} = \frac{32\pi}{3}\]

- Evaluate the Integral

The integral is the volume result from Step 5: \[\iint_S \mathbf{V}\cdot\mathbf{n}\,d\sigma = \frac{32\pi}{3}\]

Key concepts

surface integral

A surface integral extends the idea of integrating over a curve to integrating over a curved surface in three-dimensional space. Essentially, surface integrals allow us to sum up a function over a surface, much like how ordinary integrals sum up a function over an interval.

vector field

A vector field is a function that assigns a vector to every point in space. For example, in the problem, the vector field is given as \(\mathbf{V} = \left(x+x^2-y^2\right)\mathbf{i}+(2xyz-2xy)\mathbf{j}-xz^2\mathbf{k}\). Vector fields are important as they help in understanding the flow of quantities like fluid flow, heat, or even electromagnetic fields.

volume integral

Volume integrals extend the concept of integrating functions to three-dimensional space. They are used to compute quantities over a volume, such as mass, charge, or any physical quantity that can be distributed throughout a volume. In our problem, the volume integral simplifies the surface integral of a vector field using the Divergence Theorem.

divergence

Divergence is a measure of how much a vector field spreads out from a point. It is computed as the dot product of the del operator \'abla\' and the vector field \'\mathbf{V}\'. Mathematically, for a vector field \(\mathbf{V} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\), the divergence \(∇ ⋅ \mathbf{V}\) is given by \[∇ ⋅ \mathbf{V} = \frac{∂P}{∂x} + \frac{∂Q}{∂y} + \frac{∂R}{∂z}\]. In the exercise, the divergence is found to be 1, simplifying our volume integral.