Question

In Exercises 17-22, evaluate \(\int_{S} \int f(x, y, z) d S\). $$ f(x, y, z)=x^{2}+y^{2}+z^{2} ; S: z=x+2, \quad x^{2}+y^{2} \leq 1 $$

Short answer

Split the integral in step 3 and evaluate each part separately. Using definitions of integrals in polar coordinates, the final solution will be \( \frac{11π}{2} \).

Step-by-step solution

Realize the polar coordinate system

Since \( x^{2} + y^{2} \leq 1 \) is a familiar equation for a circle in 2D, convert the Cartesian coordinates to polar coordinates. Where \( x = rcos(θ) \), \( y = rsin(θ) \), and the range for \( r \) is [0, 1] and \( θ \) is [0, \( 2π \)].

Substitute and Simplify

Substitute \( z \) in \( f(x, y, z) \) by \( z = x + 2 \). Therefore, \( f(r, θ) = x^{2} + y^{2} + (x + 2)^{2} \). Plug in the polar coordinates to get \( f(r, θ) = r^{2}cos^{2}(θ) + r^{2}sin^{2}(θ) + (rcos(θ) + 2)^{2} \). Then simplify the equation to \( f(r, θ) = r^{2} + r^{2}cos^{2}(θ) + 4rcos(θ) + 4 \).

Integrate over the surface

The integral becomes \( \int_{0}^{2π} \int_{0}^{1} (r^{2} + r^{2}cos^{2}(θ) + 4rcos(θ) + 4) rdrdθ \). Evaluate the integral to obtain the solution.

Key concepts

Double Integral

The double integral is a powerful mathematical tool used to compute the volume of a two-dimensional area under a surface. The concept stems from the idea of breaking up the area into infinitesimally small rectangles and summing the volumes under the function for each rectangle. To compute a double integral, one performs an integral within an integral - this is where the term 'double' comes from.

For example, in Cartesian coordinates, a double integral of a function f(x, y) over a region R can be written as \[ \int_{R} \int f(x, y) \,dx\,dy \.\] In many cases, especially when the region R is not a simple rectangle or when dealing with polar coordinates, it can be advantageous to change the order of integration or to change the coordinate system altogether to simplify the problem.

Polar Coordinates

Polar coordinates offer an alternative to the traditional Cartesian coordinate system. In polar coordinates, each point in the plane is determined by a radius r and an angle θ (theta). The radius describes how far away the point is from the origin, while the angle indicates the direction from the origin to the point, measured from the positive x-axis.

This coordinate system is particularly useful when dealing with problems that have circular symmetry, such as circles or spheres, and when integrating over these shapes - it often simplifies calculations. For instance, in polar coordinates, the equation for a circle with radius 1 centered at the origin is simply r = 1, which is much less complex than its Cartesian counterpart x^2 + y^2 = 1.

Cartesian to Polar Conversion

Converting Cartesian coordinates (x, y) to polar coordinates (r, θ) is a straightforward process that allows for simplification when dealing with problems involving circular or spherical symmetry. The conversion formulas are given by:
\[ r = \sqrt{x^2 + y^2} \]
\[ θ = \arctan2(y, x) \]
where r is the distance from the origin to the point (x, y), and θ is the angle between the positive x-axis and the line segment connecting the origin to the point. The function arctan2 gives the angle in the correct quadrant, considering the signs of x and y.

This conversion is especially useful in our exercise, where the equation x^2 + y^2 <= 1 represents a circle. In polar coordinates, this equation simplifies to r <= 1, significantly easing the integration process.

Integrating Functions Over Surfaces

Integrating functions over surfaces extends the idea of double integrals to more complex shapes in three dimensions. A surface integral computes the 'weighted' area of a surface, where the function f(x, y, z) provides the weights for each infinitesimal area element on the surface S.

To evaluate a surface integral of the form \[ \int_S \int f(x, y, z)\, dS \.\]one often parametrizes the surface S using two parameters, typically denoted as u and v, and then expresses the differential area element dS in terms of du and dv. In the case of the given exercise, the conversion of the Cartesian equation to polar coordinates helps us to rewrite our surface element appropriately and lets us evaluate the integral in a more convenient coordinate system, leading to a simplified integration process.