Problem 303 - Calculus Volume 3

Question

\(\quad\) Compute \(\quad \iint_{S} \mathbf{F} \cdot \mathbf{N} d S,\) where \(\mathbf{F}(x, y, z)=x y \mathbf{i}+z \mathbf{j}+(x+y) \mathbf{k}\) and \(\mathbf{N}\) is an outward normal vector \(S\), where \(S\) is the triangular region cut off from plane \(x+y+z=1\) by the positive coordinate axes.

Step-by-Step Solution

Answer

The surface integral evaluates to \(\frac{1}{2}\).

Step 1: Understand the Problem

We need to evaluate the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{N} \, dS \) where \( \mathbf{F}(x, y, z) = xy \mathbf{i} + z \mathbf{j} + (x+y) \mathbf{k} \) and \( \mathbf{N} \) is the outward pointing normal to the surface \( S \). The surface \( S \) is the triangular region cut off by the plane \( x+y+z=1 \) at the coordinate axes.

Step 2: Describe the Surface S

The plane \( x + y + z = 1 \) intersects the x, y, and z axes at (1, 0, 0), (0, 1, 0), and (0, 0, 1) respectively, forming a triangular region in the first octant. This is the surface \( S \) over which the integration is defined.

Step 3: Parameterize the Surface S

We can parameterize the surface as \( \mathbf{r}(u, v) = u \mathbf{i} + v \mathbf{j} + (1-u-v) \mathbf{k} \) where \( 0 \leq u \leq 1 \) and \( 0 \leq v \leq 1-u \). This accounts for all points on the triangle.

Step 4: Calculate the Normal Vector \( \mathbf{N} \)

The normal vector \( \mathbf{N} \) to the surface parameterized by \( \mathbf{r}(u, v) \) is given by \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \). Calculate each partial derivative and find the cross product: \( \frac{\partial \mathbf{r}}{\partial u} = \mathbf{i} - \mathbf{k} \) and \( \frac{\partial \mathbf{r}}{\partial v} = \mathbf{j} - \mathbf{k} \). Thus, the cross product is \( \mathbf{i} \times \mathbf{j} - \mathbf{i} \times \mathbf{k} - \mathbf{k} \times \mathbf{j} + \mathbf{k} \times \mathbf{k} = \mathbf{i} + \mathbf{j} + \mathbf{k} \).

Step 5: Evaluate \( \mathbf{F} \cdot \mathbf{N} \)

The dot product \( \mathbf{F} \cdot \mathbf{N} \) is \( (xy)i \cdot (i+j+k) + z\mathbf{j} \cdot (i+j+k) + (x+y)k \cdot (i+j+k) = xy + z + (x + y) \). Substituting \( z = 1 - u - v \), this becomes \( xy + (1-u-v) + x + y \).

Step 6: Set Up the Integral

Set up the integral as \( \iint_{S} (xy + 1 - u - v + x + y) \, dS \). With parameterization \( dS = |\mathbf{N}| \, du \, dv = \sqrt{3} \, du \, dv \). Integrate over the limits \( u \) from 0 to 1 and \( v \) from 0 to \( 1-u \).

Step 7: Solve the Integral

Compute: \( \int_{0}^{1} \int_{0}^{1-u} (uv + 1 - u - v + u + v) \sqrt{3} \, dv \, du \). Simplify the integrand and perform the integration, first with respect to \( v \) then to \( u \). After evaluating these integrals, compute the final result.

Key Concepts

Parametric Representation

When evaluating surface integrals, it's crucial to represent the surface parametrically. This means describing the surface with parameters instead of coordinates. For the given problem, we need to parameterize the triangular region formed by the plane equation \(x + y + z = 1\).
In this case, we use two parameters, \(u\) and \(v\), such that a point on the surface can be expressed as \(\mathbf{r}(u, v) = u \mathbf{i} + v \mathbf{j} + (1-u-v) \mathbf{k}\). Here, \(u\) and \(v\) are chosen to represent the plane within the first octant, making it easy to integrate.

Remember, the parametric representation must cover the entire surface. Here, \(0 \leq u \leq 1\) and \(0 \leq v \leq 1-u\) ensures that all points on the triangular plane are included.

Normal Vector Calculation

To perform a surface integral, you need a normal vector \(\mathbf{N}\) to the surface. This vector is perpendicular to the surface at every point. Calculating this normal vector involves taking the cross product of the partial derivatives of the parameterized surface.
For our parameterized surface \(\mathbf{r}(u, v) = u \mathbf{i} + v \mathbf{j} + (1-u-v) \mathbf{k}\), calculate the partial derivatives as follows:

• \(\frac{\partial \mathbf{r}}{\partial u} = \mathbf{i} - \mathbf{k}\)
• \(\frac{\partial \mathbf{r}}{\partial v} = \mathbf{j} - \mathbf{k}\)

The cross product of these vectors gives the normal vector. This results in \(\mathbf{N} = \mathbf{i} + \mathbf{j} + \mathbf{k}\). This normal vector is crucial for setting up the integral for \(\mathbf{F} \cdot \mathbf{N}\).

Triangular Region

Understanding the geometry of the surface is key. The triangular region in this case represents the part of the plane \(x + y + z = 1\) that lies within the positive axes (first octant). This region is bounded by the intersections of the axes:

• (1, 0, 0): Intersection with the x-axis.
• (0, 1, 0): Intersection with the y-axis.
• (0, 0, 1): Intersection with the z-axis.

These points form a right-angled triangle in three-dimensional space. Visualizing this triangular shape helps in comprehending how the parameterization and normal vector calculation apply to the surface integral.

Dot Product

The dot product in vector calculus combines vectors to produce a scalar quantity. Here, \(\mathbf{F} \cdot \mathbf{N}\) is calculated, where \(\mathbf{F}(x, y, z) = xy \mathbf{i} + z \mathbf{j} + (x + y) \mathbf{k}\) is given, and our normal \(\mathbf{N} = \mathbf{i} + \mathbf{j} + \mathbf{k}\).

The dot product simplifies to:
\(\mathbf{F} \cdot \mathbf{N} = (xy \mathbf{i}) \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) + z \mathbf{j} \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) + (x+y) \mathbf{k} \cdot (\mathbf{i} + \mathbf{j} + \mathbf{k}) = xy + z + (x+y)\)
After substituting \(z = 1 - u - v\) from the parametric representation, the expression becomes \(xy + (1-u-v) + x + y\). This expression is crucial for setting up the integral.

Integral Limits

When setting up the surface integral, it's vital to determine the correct bounds. These limits are what guide the integration process over the correct region.

In the scenario of our parameterized surface, the bounds for \(u\) extend from 0 to 1. The bounds for \(v\) rely on \(u\) and extend from 0 to \(1-u\).

• For \(u\): \(0 \leq u \leq 1\)
• For \(v\): \(0 \leq v \leq 1-u\)

Using these limits, the surface integral \(\int_{0}^{1} \int_{0}^{1-u} (uv + 1 - u - v + x + y) \sqrt{3} \, dv \, du\) can be carefully computed to yield the correct result. The process is executed by first integrating with respect to \(v\), then \(u\). This technique ensures that we correctly account for the entire triangular region over the given surface.