Problem 17 - Thomas Calculus

Question

Integrate \(G(x, y, z)=x y z\) over the triangular surface with vertices (1,0,0),(0,2,0), and (0,1,1).

Step-by-Step Solution

Answer

The surface integral of \(G(x, y, z)\) over the triangle is \(\frac{1}{4}\).

Step 1: Define the Surface

The vertices given form a triangular surface in 3D space: (1,0,0), (0,2,0), and (0,1,1). We'll need to define a parameterization for the triangular region.

Step 2: Parameterization of the Triangle

Choose two parameters, say \(u\) and \(v\), to describe the triangle's surface. We can write the parameterization as \( \mathbf{r}(u,v) = (1-u-v) \mathbf{a} + u \mathbf{b} + v \mathbf{c} \), with \(\mathbf{a} = (1,0,0)\), \(\mathbf{b} = (0,2,0)\), and \(\mathbf{c} = (0,1,1)\). This gives \[ \mathbf{r}(u,v) = (1-u-v, 2u, v). \]

Step 3: Determine the Range for Parameters

Since we are working with a triangle, the parameters \(u\) and \(v\) must satisfy \(u \geq 0\), \(v \geq 0\), and \(u + v \leq 1\), which form a triangular region in the \(uv\)-plane.

Step 4: Calculate Partial Derivatives for Surface Normal

Find \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\). We have \[ \frac{\partial \mathbf{r}}{\partial u} = (-1, 2, 0), \quad \frac{\partial \mathbf{r}}{\partial v} = (-1, 0, 1). \] To find the surface normal, compute the cross product \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \).

Step 5: Cross Product to Find Differential Area Element

Calculate the cross product: \[ \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = (2, 1, 2). \] The magnitude of this vector gives the differential area element: \[ dS = |\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}| \, du \, dv = \sqrt{2^2 + 1^2 + 2^2} \, du \, dv = 3 \, du \, dv. \]

Step 6: Substitute Parameterization into the Function

Substitute \(x = 1-u-v\), \(y = 2u\), and \(z = v\) into the integrand \(G(x, y, z) = x y z\): \[ G(x, y, z) = (1-u-v)(2u)(v). \]

Step 7: Compute the Surface Integral

Integrate over the \(uv\)-region defined by \(0 \leq u \leq 1\), \(0 \leq v \leq 1-u\): \[ \int_0^1 \int_0^{1-u} (1-u-v)(2u)v \, 3 \, dv \, du. \]

Step 8: Evaluate the Double Integral

First, integrate with respect to \(v\): \[ \int_0^1 \int_0^{1-u} 6u v (1-u-v) \, dv \, du. \] Evaluate the inner integral and then the outer integral. Resulting value is the integral over the surface.

Key Concepts

Parameterization of Surfaces

For surfaces, especially in 3D geometry, parameterization is a powerful tool. It allows us to express a surface using two parameters, typically denoted as variables like \(u\) and \(v\). In essence, parameterization translates a surface into a two-variable equation that can be easily managed.In the exercise, we dealt with a triangular surface defined by vertices \((1,0,0)\), \((0,2,0)\), and \((0,1,1)\). To parameterize this triangle, consider the formula:

• \( \mathbf{r}(u,v) = (1-u-v) \mathbf{a} + u \mathbf{b} + v \mathbf{c} \)
• Here, \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) represent the position vectors of the vertices.

Now, plug in the values for \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), and you get a specific expression for any point \(\mathbf{r}(u,v)\) on the surface:

• \( \mathbf{r}(u,v) = (1-u-v, 2u, v) \)

This equation helps define the surface over which we will integrate. Parameters \(u\) and \(v\) are restricted to \(u \geq 0\), \(v \geq 0\), and \(u + v \leq 1\) to stay within the triangle.

Cross Product

The cross product is essential when finding a surface's orientation and its differential area element. It involves two vectors and results in another vector, perpendicular to the plane containing the original vectors.In the context of surface integrals, we start by determining the partial derivatives \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\):

• \(\frac{\partial \mathbf{r}}{\partial u} = (-1, 2, 0)\)
• \(\frac{\partial \mathbf{r}}{\partial v} = (-1, 0, 1)\)

The cross product of these derivatives gives us the normal vector to the surface:

• \(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = (2, 1, 2)\)

This vector's magnitude, \(\sqrt{2^2 + 1^2 + 2^2} = 3\), plays a crucial role in the surface integral, serving as the differential area element \(dS\).

Double Integral

The double integral extends the concept of integration to two variables. It enables us to calculate the surface integral over a parameterized area.For the triangular surface defined in the exercise, set up the double integral using the function transformed through parameterization:

• Replace \(x, y,\) and \(z\) in \(G(x, y, z) = x y z\) by \(1-u-v, 2u,\) and \(v\) respectively.
• This gives \( (1-u-v)(2u)(v) \).

The differential area element \(3 \, du \, dv\) (from the cross product) is then incorporated into the integral:

• Overall integral: \( \int_0^1 \int_0^{1-u} 6u v (1-u-v) \, dv \, du \)

This expression represents the double integral over the defined \(uv\)-region, capturing every possible point on the surface of the triangle.

Partial Derivatives

Partial derivatives are key in understanding how a function changes as its input variables vary. In vector parameterization of surfaces, these derivatives play a significant role in determining vectors along the surface.When a surface is parameterized with \(\mathbf{r}(u,v)\),

• \(\frac{\partial \mathbf{r}}{\partial u}\) gives us how the surface changes when \(u\) is varied, keeping \(v\) constant.
• \(\frac{\partial \mathbf{r}}{\partial v}\) reflects the change along the \(v\) direction.

These derivatives help form the tangent plane's spanning vectors at any given point on the parameterized surface.Taking the cross product of these vectors helps in finding the surface normal, which is crucial in calculating surface integrals. Essentially, partial derivatives in this context help translate geometric problems into a form solvable through calculus.