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Evaluate \(\int_{S} \int f(x, y) d S\). \(f(x, y)=x+y\) \(S: \mathbf{r}(u, v)=2 \cos u \mathbf{i}+2 \sin u \mathbf{j}+v \mathbf{k}\) \(\quad 0 \leq u \leq \frac{\pi}{2}, \quad 0 \leq v \leq 2\)

Short Answer

Expert verified
\(\int_S f dS = 16\)

Step by step solution

01

Substitution of Variables

From the parametric equation, we can see that \(x = 2 \cos{u}\) and \(y = 2 \sin{u}\). Therefore, we get \(f(u,v) = 2(\cos{u} + \sin{u})\)
02

Differential Transformation

Calculate the differential \(dS\) over the surface. We first need \(| \mathbf{r}_u \times \mathbf{r}_v |\). So, take derivatives: \(\mathbf{r}_u = -2 \sin u \mathbf{i} + 2 \cos u \mathbf{j} + 0 \mathbf{k}\) and \(\mathbf{r}_v = 0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k}\). Do their cross product, which results in \(\mathbf{r}_u \times \mathbf{r}_v = 2 \cos u \mathbf{i} + 2 \sin u \mathbf{j}\). The magnitude of their cross product is then \(2\). This means that \(dS = 2dudv\)
03

Evaluate the Integral

Substitute \(f(u,v)\) and \(dS\) into \( \int_S f dS\), and perform the double integral: \(\int_0^2 \int_0^{ \pi/2} 2(2(\cos{u} + \sin{u})) du dv = 4 \int_0^2 \int_0^{ \pi/2} (\cos{u} + \sin{u}) du dv = 4 \int_0^2 ([\sin{u} - \cos{u}]_0^{ \pi/2} ) dv = 4 \int_0^2 (1+1) dv = 4 \int_0^2 2 dv = 16\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Integral
The double integral is a crucial concept in calculus when dealing with functions of two variables over a two-dimensional region. In terms of computing areas and volumes, the double integral allows us to extend the idea of a single integral over one dimension to calculate more complex quantities. The process involves integrating a function, typically denoted as \(f(x,y)\), over a region in the plane.

Imagine summing up an infinite number of infinitesimally small rectangles, each contributing a tiny area to the total. In our exercise, the double integral finds the

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