Question

In Exercises 1-4, evaluate \(\int_{S} \int(x-2 y+z) d S\). $$ S: z=4-x, \quad 0 \leq x \leq 4, \quad 0 \leq y \leq 4 $$

Short answer

The double integral of the vector field over the surface S is 0.

Step-by-step solution

Establish the Integral

The double integral on S can be written as: \(\int_{0}^{4} \int_{0}^{4} (x - 2y + 4 - x) dy dx\). Here, \(z\) was replaced with \(4 - x\) from the equation of the surface.

Simplify the Integrals

Simplify the integral, it will lead to a simpler integral to integrate: \(\int_{0}^{4} \int_{0}^{4} (-2y + 4) dy dx\)

Compute the Integral with respect to y

Perform the inside integral first with respect to \(y\). This will lead to: \(-2 \int_{0}^{4} [y^2 / 2 ]_{0}^{4} + 4 \int_{0}^{4} y dy dx = \int_{0}^{4}(-16+16)dx = 0\) since the term \(-2y^2\) from 0 to 4 evaluates to -16 and the term \(4y\) from 0 to 4 evaluates to 16.

Compute the Integral with respect to x

Carry out the outside integral: \(\int_{0}^{4}0 dx = 0\)

Key concepts

Surface Integral

The concept of a surface integral is similar to that of a line integral, but instead of integrating over a curve, we are integrating over a surface in space. In the exercise, the surface integral \(\int_{S} \int(x-2 y+z) d S\) is considered for a function over a given surface S. A surface integral computes the flow of a vector field across a surface, or, as in this case, it can also integrate a scalar field over a patch of surface.

In order to evaluate this type of integral, we must first understand the surface over which we are integrating. Here, the surface S is defined by the equation \(z=4-x\), with x and y boundaries from 0 to 4. To compute the integral, we must express \(z\) in terms of \(x\) and \(y\), and then project the surface onto the x-y plane, which simplifies the evaluation process.

Integral Simplification

When faced with a complex integral, simplifying the expression can make evaluation much easier. In the provided step-by-step solution, simplification is demonstrated when the original integral \(\int_{0}^{4} \int_{0}^{4} (x - 2y + 4 - x) dy dx\) is made simpler by canceling out the terms \(x\) and \(-x\).

After simplification, the double integral becomes \(\int_{0}^{4} \int_{0}^{4} (-2y + 4) dy dx\), which is much more manageable. Simplification can also involve algebraic manipulation, substitution, or even changing the order of integration, which can facilitate easier integration and reduce errors during calculation.

It's important to identify opportunities to simplify an integral before proceeding to compute it. Eliminating terms or factors that can cancel each other out can greatly reduce the complexity of the integral, thus reducing the computational burden.

Integral Evaluation

Evaluating an integral involves finding the accumulation of a quantity described by the function being integrated. In the context of double integrals, evaluation is generally performed by integrating the function with respect to each variable sequentially, as illustrated in steps 3 and 4 of the solution.

The process begins with integrating the innermost variable, which is often represented by an inner integral. In this case, integration with respect to \(y\) yields \(\int_{0}^{4}(-16+16)dx = 0\) because the functions \(y^2 / 2\) and \(y\) evaluated from 0 to 4 are offsetting each other.

After the interior integral is evaluated, you proceed to the next integral. In this problem, this is a trivial step since integrating zero over any interval will result in zero. Hence, the evaluation of the integral with respect to \(x\) did not change the already obtained value. Integral evaluation is crucial as it gives the final accumulative result of the given function over the specified domain.