Question

In Exercises 13-16, evaluate \(\int_{S} \int f(x, y) d S\). $$ \begin{array}{l} f(x, y)=x+y ; S: \mathbf{r}(u, v)=4 u \cos v \mathbf{i}+4 u \sin v \mathbf{j}+3 u \mathbf{k} \\ 0 \leq u \leq 4, \quad 0 \leq v \leq \pi \end{array} $$

Short answer

The value of the double integral is the final answer obtained after evaluating the definite integral in Step 3

Step-by-step solution

Write the Given

The function \(f(x, y) = x + y\), and the parametric equation of the surface S is \(\mathbf{r}(u, v) = 4u \cos v \mathbf{i} + 4u \sin v \mathbf{j} + 3u \mathbf{k}\), with \(0 \leq u \leq 4\), \(0 \leq v \leq \pi\).

Compute the Integral Expressions

Under the change of variables, obtain the expression for \(f(x, y)\) in terms of \(u, v\) by substituting \(x = 4u \cos v\) and \(y = 4u \sin v\) into \(f(x, y) = x + y\). Now, compute the magnitude of the cross product of the partial derivatives of the vector function \(\mathbf{r}(u, v)\) with respect to \(u\) and \(v\), which is the area element \(dS\) on the surface S in terms of \(du dv\). The integral to be evaluated becomes \(\int_{S} \int f(u, v) | \mathbf{r}_{u} \times \mathbf{r}_{v} | du dv\).

Evaluate the Integral

Use a mathematical software or manual calculations to compute the double integral obtained in the previous step. Make sure the definite integration rules are followed with respect to \(u\) and \(v\).

Key concepts

Double Integral

The double integral is a powerful tool in calculus, particularly when we want to calculate the volume under a surface or the area of a curved surface. It extends the concept of a single integral to two dimensions, where we are now integrating over a region in the xy-plane.

Mathematically, the double integral of a function f(x, y) over a region R is denoted by \[\int \int_R f(x, y) \, dA\], where dA represents an infinitesimal element of area within the region R. To compute a double integral, one often iterates the process of integration: by first integrating with respect to y and then with respect to x, or vice versa; the order of integration depends on the limits of integration for the particular region.

In our exercise, the double integral will be used to find the quantity \[\int_{S} \int f(x, y) \, dS\], which is computed over a surface S rather than a flat region R. This involves transforming the function into a form that can be integrated over the parameters that define the surface.

Parametric Surfaces

In calculus, parametric surfaces provide a way to describe a surface in three-dimensional space using two parameters. This is akin to how a parametric line in two dimensions uses a single parameter.

A parametric surface S is given by a vector function \[\mathbf{r}(u, v) = x(u, v)\mathbf{i} + y(u, v)\mathbf{j} + z(u, v)\mathbf{k}\], where \(u, v\) are parameters that 'trace out' the surface as they vary within their defined ranges. For each pair \(u, v\), there corresponds a point \[\mathbf{r}(u, v)\] on the surface. In the given exercise, the surface S is defined by the vector function \[\mathbf{r}(u, v)=4u \cos v \mathbf{i}+4u \sin v \mathbf{j}+3u \mathbf{k}\] with the parameters \(u\) and \(v\) within specified intervals.

Parametric surfaces allow us to calculate surface integrals by transforming them into double integrals over the parameter domain, making it possible to handle surfaces that cannot be represented as functions of two variables in the form z = f(x, y).

Cross Product

The cross product is an operation on two vectors in three-dimensional space that results in a new vector which is perpendicular to the plane containing the original vectors. Given two vectors \(\mathbf{a}\) and \(\mathbf{b}\), their cross product \(\mathbf{a} \times \mathbf{b}\) has a magnitude equal to the area of the parallelogram spanned by \(\mathbf{a}\) and \(\mathbf{b}\).

The magnitude of the cross product is leveraged in surface integrals to find an area element \(dS\), which represents the infinitesimal 'patch' of surface area. In our exercise, we compute \(\mathbf{r}_u \times \mathbf{r}_v\), the cross product of the partial derivatives of \(\mathbf{r}(u, v)\) with respect to \(u\) and \(v\), to obtain the normal vector that defines the orientation of the surface. Its magnitude \[|\mathbf{r}_u \times \mathbf{r}_v|\] is the area element \(dS\).

When we integrate \(f(x, y)\) over the surface S using the expression for \(dS\), we can account for the 'stretching' of area due to the surface's curvature and orientation in space, thus turning our surface integral into a double integral over the parameters of the surface.