Problem 292 - Calculus Volume 3

Question

Evaluate \(\iint_{S}(x+y+z) d \mathbf{S},\) where \(S\) is the surface defined \(\quad\) parametrically \(\quad\) by \(\mathbf{R}(u, v)=(2 u+v) \mathbf{i}+(u-2 v) \mathbf{j}+(u+3 v) \mathbf{k}\) for \(0 \leq u \leq 1,\) and \(0 \leq v \leq 2\)

Step-by-Step Solution

Answer

The surface integral evaluates to -40.

Step 1: Identify the Surface Parametrization

The surface \( S \) is defined by the parametrization \( \mathbf{R}(u, v) = (2u + v) \mathbf{i} + (u - 2v) \mathbf{j} + (u + 3v) \mathbf{k} \). This means \( x = 2u + v \), \( y = u - 2v \), and \( z = u + 3v \). The parameters \( u \) range from 0 to 1 and \( v \) range from 0 to 2. This will help define the integration limits later.

Step 2: Compute Surface Element dS

To compute the surface element \( d\mathbf{S} \), first find the partial derivatives \( \frac{\partial \mathbf{R}}{\partial u} = 2 \mathbf{i} + \mathbf{j} + \mathbf{k} \) and \( \frac{\partial \mathbf{R}}{\partial v} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \). Next, find the cross product \( \frac{\partial \mathbf{R}}{\partial u} \times \frac{\partial \mathbf{R}}{\partial v} \), which gives the normal vector to the surface. Calculating the cross product, we obtain the normal vector \( \mathbf{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 1 \ 1 & -2 & 3 \end{vmatrix} = 5 \mathbf{i} - 5 \mathbf{j} - 5 \mathbf{k} \). Thus, \( d\mathbf{S} = \mathbf{N} \, du \, dv = (5 \mathbf{i} - 5 \mathbf{j} - 5 \mathbf{k}) \, du \, dv \).

Step 3: Set Up the Double Integral

The next step involves setting up the integral \( \iint_{S} (x + y + z) \, d\mathbf{S} \). Substitute \( x = 2u + v \), \( y = u - 2v \), and \( z = u + 3v \) into the integrand: \( x + y + z = (2u + v) + (u - 2v) + (u + 3v) = 4u + 2v \). So the integral becomes: \( \iint (4u + 2v) \cdot (5 \mathbf{i} - 5 \mathbf{j} - 5 \mathbf{k}) \, du \, dv \). Expand this expression to integrate each component of \( d\mathbf{S} \) over each parameter.

Step 4: Simplify and Evaluate the Integral

Since \( d\mathbf{S} = (5 \mathbf{i} - 5 \mathbf{j} - 5 \mathbf{k}) \, du \, dv \) is a vector, we multiply the scalar \( 4u + 2v \) by the normal vector. We focus on the scalar parts for the integral: \[ \int_0^1 \int_0^2 (4u + 2v)(-5) \ du \ dv = -5 \int_0^1 \left( \int_0^2 (4u + 2v) \, dv \right) du \].Compute the inner integral: \( \int_0^2 (4u + 2v) \ dv = [4uv + v^2]_0^2 = 8u + 4 \). The integral becomes \( -5 \int_0^1 (8u + 4) \, du \). The final integral evaluates to \( -5 [4u^2 + 4u]_0^1 = -5 [(4(1)^2 + 4(1)) - (0)] = -40 \).

Step 5: Interpret the Result

The integral evaluates to \(-40\). This result accounts for the orientation of the surface, determined by the direction of the normal vector found through the cross product. The negative value indicates the integration over the oriented surface considering the downward pointing normal vector.

Key Concepts

Parametrized Surfaces

In vector calculus, a parametrized surface is a smooth mapping that represents a surface in three-dimensional space using two parameters. Imagine a piece of fabric that can stretch and bend; parametrization is like giving coordinates to each point on the fabric using variables like \( u \) and \( v \).
For our surface \( S \), we use the parametrization \( \mathbf{R}(u, v) = (2u + v) \mathbf{i} + (u - 2v) \mathbf{j} + (u + 3v) \mathbf{k} \). This means that:

• \( x = 2u + v \)
• \( y = u - 2v \)
• \( z = u + 3v \)

These expressions connect the parameters \( u \) and \( v \) to points \( (x, y, z) \) on the surface.

This process is useful in defining the domain for integration, allowing us to move from the messy three-dimensional space to a clean two-dimensional parameter space, simplifying calculations.

Cross Product

The cross product is a fundamental operation in vector calculus, especially when dealing with surfaces and orientations.
When two vectors in space are crossed, they produce a third vector that is perpendicular to the plane containing the original vectors. This is crucial for calculating the direction of a surface's orientation, which is reflected in the normal vector.Calculating the cross product of the partial derivatives \( \frac{\partial \mathbf{R}}{\partial u} \) and \( \frac{\partial \mathbf{R}}{\partial v} \) gives us the normal vector \( \mathbf{N} \) to the surface:

• \( \frac{\partial \mathbf{R}}{\partial u} = 2 \mathbf{i} + \mathbf{j} + \mathbf{k} \)
• \( \frac{\partial \mathbf{R}}{\partial v} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \)

The cross product results in:\[ \mathbf{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 1 \ 1 & -2 & 3 \end{vmatrix} = 5 \mathbf{i} - 5 \mathbf{j} - 5 \mathbf{k} \]This vector \( \mathbf{N} \), also known as the normal vector, determines how the surface is oriented in space.

Vector Calculus

Vector calculus is the field of mathematics that deals with vectors and their operations, crucial for understanding surface integrals.
It provides tools to investigate vector fields and perform complex calculations like determining the flux through a surface or the orientation of vector fields on surfaces. In our exercise, vector calculus techniques help define the surface \( S \) using parametrization and compute normal vectors, which are key to setting up and evaluating integrals on surfaces.
A surface integral, like the one in our example, uses vector calculus to sum up quantities over a surface, bridging scalar functions with geometry. This integral sums the function \( x+y+z \) over the surface defined by our parameters \( u \) and \( v \). With the help of vector calculus, we translate complex geometric configurations into tractable integrals.

Double Integral

Double integrals are an extension of single-variable integration into two dimensions, allowing us to integrate over areas or surfaces.
In our case, the surface integral \( \iint_{S} (x + y + z) \, d\mathbf{S} \) involves a double integral because we are integrating over a surface defined by two parameters, \( u \) and \( v \).
The steps involve:

• Substituting \( x = 2u + v \), \( y = u - 2v \), \( z = u + 3v \) into the function for integration.
• Using the cross product to find \( d\mathbf{S} = (5 \mathbf{i} - 5 \mathbf{j} - 5 \mathbf{k}) \, du \, dv \), which acts as our differential area element on \( S \).
• Setting up and solving the integral: \[ -5 \int_0^1 \left( \int_0^2 (4u + 2v) \, dv \right) du \]

This process uses techniques of double integration to "sum" a function over a chosen area or surface, providing insight into the cumulative effect of a function over that space.