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 \(R=\\{(x, y):-1 \leq x \leq 3,0 \leq y \leq 2\\}\) \(f(x, y)=24-3 x-4 y\)

Short answer

Answer: The volume of the solid is 109 cubic units.

Step-by-step solution

1. Set Up Double Integral

To set up the double integral, we will integrate the function \(f(x, y) = 24 - 3x - 4y\) over the region \(R\) with respect to the variables \(x\) and \(y\). The limits for the variable \(x\) are \(-1 \leq x \leq 3\), and the limits for the variable \(y\) are \(0 \leq y \leq 2\). We will first integrate with respect to \(x\), and then integrate with respect to \(y\): $$ \int_{0}^{2} \int_{-1}^{3} (24 - 3x - 4y) dx dy $$

2. First Integration (with respect to x)

Now, we will integrate the function \(24 - 3x - 4y\) with respect to \(x\): $$ \int_{0}^{2} \left[ 24x - \frac{3}{2} x^2 - 4xy \right]_{-1}^{3} dy $$

3. Evaluate the bounds for x

Plug in the bounds for \(x\): $$ \int_{0}^{2} \left[ (72 - 27 - 12y) - (-24 + \frac{3}{2} + 4y) \right] dy $$ Simplify the expression: $$ \int_{0}^{2} (96 - \frac{51}{2} - 16y) dy $$

4. Second Integration (with respect to y)

Now, we will integrate the function \(96 - \frac{51}{2} - 16y\) with respect to \(y\): $$ \left[ 96y - \frac{51}{2}y - 8y^2 \right]_{0}^{2} $$

5. Evaluate the bounds for y

Plug in the bounds for \(y\): $$ (192 - 51 - 32) - (0) $$ Simplify the expression: $$ Volume = 141 - 32 = 109 $$ The volume of the solid beneath the plane \(f(x, y) = 24 - 3x - 4y\) and above the region \(R=\\{(x, y):-1 \leq x \leq 3,0 \leq y \leq 2\\}\) is 109 cubic units.

Key concepts

Double Integral Volume Calculation

Understanding how to calculate the volume of a solid using double integrals is an essential concept in calculus. The process involves setting up an integral that represents the volume of the solid above a particular region and beneath a surface in the xyz-space. The double integral effectively sums up the volumes of infinitesimally thin slices of the solid.

To calculate the volume using a double integral, we consider a surface defined by some function f(x, y) above a region R in the xy-plane. The volume V is given by the integral \[ V = \int \!\int_{R} f(x, y) \, dx \, dy \]. In other words, we integrate the function f(x, y) over the area of R. To do this, we choose an order of integration, which is often determined by the region's bounds or the integrand's simplicity after integrating with respect to x or y.

In the given exercise, we are finding the volume of a solid beneath the plane f(x, y) = 24 - 3x - 4y and above the region R. The procedure begins by integrating the function with respect to x, followed by integrating the result with respect to y. It is crucial to correctly set the limits of integration based on the given region R.

Integration with Respect to X and Y

When performing a double integral to calculate volume, we integrate the function with respect to both x and y.

Integrating with Respect to X

The specific order in which we perform these integrations can vary, but for our example, the function f(x, y) = 24 - 3x - 4y is first integrated with respect to x, the horizontal component. We treat y as a constant during this step and integrate across the x-bounds of region R.

Integrating with Respect to Y

Next, we integrate the resulting expression, now a function of y only, over the y-bounds of region R. This step involves treating x as having been 'eliminated' from the integral, reflecting the accumulation of volume in the vertical direction. This layer-by-layer approach helps us build the volume from the ground up.

Evaluating Bounds in Integrals

Evaluating the bounds in integrals is a systematic process of replacing the variable of integration with the limits of integration. It's essential for calculating areas or volumes.

For the given exercise, after integrating with respect to x, we evaluate at the upper and lower bounds for x and subtract the results. This gives us an expression solely in terms of y. We do something similar after integrating with respect to y; evaluating at the upper and lower bounds for y gives us the final numerical volume of the solid. Correctly evaluating these bounds is critical since it represents the 'actual' volume by accounting for the solid's limits in each direction.

Solid of Revolution

The concept of a solid of revolution is a special application of integration used to find the volume of an object created by rotating a shape around an axis. Although our specific exercise does not involve a solid of revolution, understanding this concept is beneficial for visualizing volume calculations in calculus.

We imagine a two-dimensional area, often under a curve or between curves, rotating about an axis. This rotation generates a three-dimensional solid. To calculate the volume of a solid of revolution, we would typically use the disk, washer, or shell method, depending on the nature of the region and the axis of rotation. These methods involve setting up an integral that sums up the volumes of circular disks, washers, or cylindrical shells, akin to how double integrals sum up elements of volume in a non-revolving context.