Question

Diamond region Consider the region \(R=\\{(x, y):|x|+|y| \leq 1\\}\) shown in the figure. a. Use a double integral to show that the area of \(R\) is 2 b. Find the volume of the square column whose base is \(R\) and whose upper surface is \(z=12-3 x-4 y\) c. Find the volume of the solid above \(R\) and beneath the cylinder \(x^{2}+z^{2}=1\) d. Find the volume of the pyramid whose base is \(R\) and whose vertex is on the \(z\) -axis at (0,0,6)

Short answer

a. By evaluating the double integral (area of R), $$ \int\limits_{-1}^1 (2-2x ) dx = \Bigl[ -2x^2 +2x \Bigr]_{-1}^1 = 2 $$ The area of region \(R\) is equal to 2. b. To find the volume of the square column, evaluate the double integral: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} (12-3x-4y) dy\ dx = \int\limits_{-1}^1 (12-3x) dx = \Bigl[ 12x -1.5 x^2 \Bigr]_{-1}^1 = 12 $$ The volume of the square column is 12. c. To find the volume of the solid beneath the cylinder, evaluate the double integral: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} \sqrt{1-x^2} dy\ dx = \pi $$ The volume of the solid beneath the cylinder is \(\pi\). d. To find the volume of the pyramid, evaluate the double integral: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} (6-6|x|-6|y|) dy\ dx = \frac{1}{2} $$ The volume of the pyramid is \(\frac{1}{2}\).

Step-by-step solution

a. Finding the area of region \(R\) using double integral

To find the area of region \(R\) using a double integral, we first need to find limits of integration for \(x\) and \(y\). Note that \(R=\{(x, y):|x|+|y| \leq 1\}\). An appropriate Cartesian limits is \(-1\le x\le1\) and \(|-1-x|\le y\le 1-x\). Then, the double integral to find the area of \(R\) will be: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} dy\ dx $$ Since the integrand for area is 1, we can compute the integral: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} dy\ dx = \int\limits_{-1}^1 (1-x - (-1-x)) dx $$ After evaluating the integral, we can verify that the area of \(R\) is equal to 2.

b. Finding the volume of the square column

To find the volume of the square column whose base is \(R\) and whose upper surface is \(z=12-3x-4y\), we need to set the integrand to \(z=12-3x-4y\). Using the same limits of integration as in part a, the double integral representing the volume can be written as: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} (12-3x-4y) dy\ dx $$ We will evaluate this integral to find the volume of the square column.

c. Finding the volume of the solid beneath the cylinder

We want to find the volume of the solid above \(R\) and beneath the cylinder \(x^2+z^2=1\). To do this, we need to express the equation for the cylinder as \(z = \sqrt{1-x^2}\). The double integral for the volume will then be: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} \sqrt{1-x^2} dy\ dx $$ We will evaluate this integral to find the volume of the solid beneath the cylinder.

d. Finding the volume of the pyramid

To find the volume of the pyramid whose base is \(R\) and whose vertex is on the \(z\)-axis at (0,0,6), we can use a similar approach as in the previous parts. First, we need to express the upper surface of the pyramid as \(z = 6-6|x|-6|y|\). Then, the double integral for the volume of the pyramid will be: $$ \int\limits_{-1}^1 \int\limits_{-1-x}^{1-x} (6-6|x|-6|y|) dy\ dx $$ We will evaluate this integral to find the volume of the pyramid.

Key concepts

Area Calculation

In mathematics, calculating the area of a region can be efficiently handled through the concept of double integrals. This is especially useful when the region has irregular shapes or is defined by particular inequalities. For our region \( R = \{(x, y): |x|+|y| \leq 1\} \), we observe that it resembles a diamond-shaped area on the xy-plane. This diamond shape can be bounded using Cartesian coordinates with limits \(-1 \leq x \leq 1\) and \(-1 - x \leq y \leq 1 - x\).

Using a double integral with the integrand as 1, since we're only interested in the area, we evaluate the following integral:

\[ \int_{-1}^1 \int_{-1-x}^{1-x} 1 \ dy \ dx \]

Breaking it down, this double integral essentially sums up small rectangles within the diamond region until the total area is calculated. After working through the integration, the resultant area of the region \( R \) is found to be 2. This establishes a foundation in using double integrals for area calculation, allowing us to handle more complex regions and shapes in calculus.

Volume Calculation

Volume calculation using calculus extends the principles of integration to the third dimension. In exercise step (b), we compute the volume of a three-dimensional shape, specifically a square column above the region \( R \). The upper surface of this column is defined by \( z = 12 - 3x - 4y \).

To find the volume, we express it through a double integral:

\[ \int_{-1}^1 \int_{-1-x}^{1-x} (12 - 3x - 4y) \, dy \, dx \]

Here, the integrand \( (12 - 3x - 4y) \) represents the height of the column at any point \((x, y)\) within the base region \( R \). By integrating this over the base region, we add up all the contributions of these heights to find the total volume. This evaluation gives us a systematic way to deal with volumes bounded by non-uniform upper surfaces, showcasing the power of double integrals in volume calculations.

Cylindrical Coordinates

Cylindrical coordinates provide a valuable perspective in calculus, especially when dealing with volume problems involving rotational symmetry. For our specific example, finding the volume of a solid beneath the cylinder \( x^2 + z^2 = 1 \), the conversion to cylindrical coordinates can simplify the equation.

The cylinder equation can be rewritten as \( z = \sqrt{1 - x^2} \). Using cylindrical coordinates is particularly helpful when our shapes are symmetrical around one of the coordinate axes. It turns out, however, in this particular exercise, the region \( R \) is more conveniently handled in Cartesian rather than cylindrical forms due to its diamond shape. Despite this, knowledge of both coordinate systems enhances flexibility in tackling complex integrals.

Cylindrical coordinates can be especially advantageous in problems involving volumes of rotational bodies or when simplifying the computation of certain limits. However, in this problem, sticking to Cartesian simplifies handling the region \( R \). By keeping this versatility in mind, you can always revert to the system that makes integration more manageable, depending on the problem structure.

Pyramids in Calculus

Calculating the volume of pyramids using calculus reveals how integration can be adapted to find volumes of intricate shapes. Specifically, pyramids feature a vertex and a polygonal base that can assume various forms. Here, we're tasked with finding the volume of a pyramid with a base of region \( R \) and a vertex precisely at the point \((0, 0, 6)\).

For such a pyramid, the volume can be represented by a double integral incorporating the height expression derived from its vertex. The function \( z = 6 - 6|x| - 6|y| \) describes the side slopes extending from the vertex to the base corners. Using this within our integral,\[ \int_{-1}^1 \int_{-1-x}^{1-x} (6 - 6|x| - 6|y|) \, dy \, dx \]

allows us to stack up infinitesimal volume elements from base to vertex.

Integrating over the defined base region and applying limits for \(x\) and \(y\), we closely analyze how each point within the pyramid contributes to its total volume. Calculus enables us to break down complex geometrical entities into manageable computations, facilitating precise volume determinations for not only pyramids but a vast array of 3D structures.