Question

Find the volume of the following solids. The solid beneath the plane \(f(x, y)=24-3 x-4 y\) and above the region \(R=\\{(x, y):-1 \leq x \leq 3,0 \leq y \leq 2\\}\)

Short answer

Answer: The volume of the solid is 136 cubic units.

Step-by-step solution

Identify the bounds of integration

First, we'll identify the bounds of integration for the double integral using the given region R: - The x-bounds are: \(-1 \leq x \leq 3\) - The y-bounds are: \(0 \leq y \leq 2\)

Set up the double integral

Now set up the double integral using the bounds of integration and the given function \(f(x, y) = 24 - 3x - 4y\): $$V = \int_{-1}^3 \int_{0}^2 (24 - 3x - 4y) dy\, dx$$ Notice that this is an iterated integral: we will first integrate with respect to y, and then with respect to x.

Integrate with respect to y

Integrate the inner integral with respect to y: $$\int_0^2 (24 - 3x - 4y) dy = [24y - 3xy - 2y^2]_0^2$$ $$= (48 - 6x - 8) - (0) = 40 - 6x$$

Integrate with respect to x

Now, integrate the outer integral with respect to x using our result from step 3: $$V = \int_{-1}^3 (40 - 6x) dx = [40x - 3x^2]_{-1}^3$$ $$= (120 - 27) - (-40 + 3)$$ $$= 93 + 43 = 136$$ The volume of the solid beneath the plane and above the rectangular region is 136 cubic units.

Key concepts

Iterated Integral

An iterated integral is a key tool in evaluating double integrals and finding the volume of a solid over a given region. When working with functions of two variables, we often use iterated integrals to perform integration successively with respect to each variable. For our problem, first, you integrate with respect to one variable (typically y), while treating the other variable as a constant.
Next, you integrate the resulting expression with respect to the second variable (x in our case).

This stepwise approach allows us to compute the double integral over a specified region, which is essential in calculating the volume beneath a surface defined by a function.

Volume of a Solid

Finding the volume of a solid in multivariable calculus involves integrating a function over a defined region in the xy-plane. Here, the function \( f(x, y) = 24 - 3x - 4y \) represents a plane in 3-dimensional space. Our goal is to determine the volume that lies beneath this plane and above the rectangular region, R.

By setting up a double integral \( V = \int_{-1}^3 \int_{0}^2 (24 - 3x - 4y) \, dy \, dx \), we can "stack up" layers of infinitesimally small volume elements to encompass the entire specified region. Each element contributes to the total volume, resulting in a calculated value of 136 cubic units.

Bounds of Integration

The bounds of integration define the region over which the function will be integrated. For this exercise, the solid's base is constrained by the rectangular region \( R = \{(x, y):-1 \leq x \leq 3, 0 \leq y \leq 2\} \).

These bounds dictate the limits of our iterated integral, with \( x \) varying from \(-1\) to \(3\), and \( y \) from \(0\) to \(2\).
These constraints fully describe the region under the plane where the volume will be calculated, ensuring that only the specified area is taken into account during integration.

Multivariable Calculus

Multivariable calculus extends ordinary calculus to functions of several variables. It is used to study and solve problems involving surfaces and volumes in higher dimensions. Understanding how to work with multiple variables is essential when dealing with physical models and real-world applications that extend beyond two-dimensional spaces.

In this context, our problem involves calculating the volume of a 3D region under a plane and above a specific region in the xy-plane. Multivariable calculus provides the tools and techniques needed to handle such analyses by employing partial derivatives and multiple integrals to explore changes and areas in different planes.