Problem 69 - Calculus Early Transcendentals

Question

Find the volume of the following solid regions. The solid above the region \(R=\\{(x, y): 0 \leq x \leq 1\), \(0 \leq y \leq 1-x\\}\) bounded by the paraboloids \(z=x^{2}+y^{2}\) and \(z=2-x^{2}-y^{2},\) and the coordinate planes in the first octant

Step-by-Step Solution

Answer

Answer: The volume of the solid region is 2 cubic units.

Step 1: Identify the functions to be subtracted

Since we want to find the volume between two paraboloid functions, we will subtract one function from the other. Here, the paraboloid functions are given by \(z_1 = x^2+y^2\) and \(z_2 = 2-x^2-y^2\). As we're looking for the volume of the solid above \(z_1\) and below \(z_2\), we need to subtract \(z_1\) from \(z_2\).

Step 2: Set up the integral

Using the Cartesian coordinate system, the volume of the solid region can be represented by the following double integral: $$V = \iint_R (z_2 - z_1) \, dA$$ Where \(R = \{(x, y): 0 \leq x \leq 1\), \(0 \leq y \leq 1-x\}\), and \(dA = dxdy\).

Step 3: Plug in the functions and set the limits

Replace \(z_2 - z_1\) with the respective functions and set the limits of integration according to the specified region R: $$V = \int_{0}^{1} \int_{0}^{1-x} (2-x^2-y^2 - (x^2+y^2)) \, dy \, dx$$

Step 4: Simplify the integrand

Simplify the expression inside the integral: $$V = \int_{0}^{1} \int_{0}^{1-x} (2-2x^2-2y^2) \, dy \, dx$$

Step 5: Evaluate the inner integral

Evaluate the integral with respect to y: $$V = \int_{0}^{1} \left[-\frac{4}{3}y^3 + 2(1-x^2)y\right]_{0}^{1-x} dx$$

Step 6: Plug in the limits of integration for y

Replace y with the limits of integration: $$V = \int_{0}^{1} \left[-\frac{4}{3}(1-x)^3 + 2(1-x^2)(1-x)\right] dx$$

Step 7: Simplify the integrand

Simplify the expression inside the integral: $$V = \int_{0}^{1} \left(-\frac{4}{3}(1-3x+3x^2-x^3) + 2(1-x^2)(1-x)\right) dx$$

Step 8: Evaluate the integral with respect to x

Evaluate the integral and find the volume of the solid region: $$V = \left[-\frac{1}{3}x^4+x^3-\frac{7}{6}x^2+\frac{11}{9}x\right]_{0}^{1}$$

Step 9: Plug in the limits of integration for x

Replace x with the limits of integration: $$V = \left[-\frac{1}{3}(1)^4+(1)^3-\frac{7}{6}(1)^2+\frac{11}{9}(1)\right] - \left[0\right]$$

Step 10: Simplify the expression

Simplify the final expression to obtain the volume of the solid region: $$V = -\frac{1}{3} + 1 - \frac{7}{6} + \frac{11}{9} = \frac{9}{9} - \frac{3}{9} - \frac{21}{9} + \frac{33}{9} = \frac{18}{9}$$ The volume of the solid region is \(\frac{18}{9} = 2\) cubic units.

Key Concepts

Double Integrals

Double integrals are a powerful tool used in calculus to determine volumes under surfaces, among other applications. In simpler terms, when you calculate a double integral, you're essentially finding the "total weight" of a surface if each tiny bit has a constant "density" or value. Double integrals help in computing areas, volumes, and other quantities in two-dimensional regions.
Let's break down the double integral process:

• First, you set up your region of interest, which is usually a part of the coordinate plane. This region dictates the limits of your integral.
• Then, a function defines the surface above this region. In our case, the function related to the paraboloids helps define boundaries for computing volume.
• The process involves integrating the "height" or "difference of heights" of your functions across this region.

When you see a typical double integral, such as \( \iint_R f(x, y) \, dA \), it involves two integrations. You integrate firstly over one variable (say \(y\)) while keeping \(x\) constant, then over the other variable. This step-by-step integration helps in uncovering the three-dimensional realm hidden in two-variable functions.

Paraboloid

A paraboloid is a symmetrical, bowl-shaped surface that looks like an umbrella turned inside out. It can be described by a quadratic equation involving two variables. For instance, in this exercise, we used two paraboloid surfaces that confined a solid below and above:

• \(z_1 = x^2 + y^2\) is a paraboloid stretching upwards, funneling away from the origin point. This forms a depression at any point above the region \(R\).
• \(z_2 = 2 - x^2 - y^2\) is an inverted paraboloid, bending downward, aimed at surrounding space with arms reaching out.

When two such paraboloids intersect, they form a closed, finite region above a specific area—here calculated using double integrals. The trick is to find the intersecting volume precisely, as these quadratic forms curve gently around each other, creating interesting shapes and challenges in geometry.

Coordinate Planes

In mathematics, the coordinate plane helps us locate points using a pair of numerical values called coordinates. For volume calculation, specifically in this exercise, we deal with regions in the 'first octant' on the three-dimensional coordinate system. This is the section of space where all of \(x\), \(y\), and \(z\) are non-negative.
Let's look at how coordinate planes help structure our calculations:

• The \(x-y\) coordinate plane acts as the bedrock upon which regions and equations operate. Points on this plane are where we evaluate differences in height for our double integrals.
• The 'first octant' conditions tell us that the solid is confined strictly to where \(x \geq 0, y \geq 0, \) and \(z \geq 0\). Essentially, it means focusing where both positive coordinates make calculations neater and relevant.
• Boundaries like \(x = 0\) and \(y = 0\) are what confine our region \(R\), guiding the setup of your double integral limits.

Using coordinate planes not only grounds calculations but provides visual clarity to how geometric volumes operate, perfectly aligning algebraic functions with physical manifested shapes like paraboloids.