Question

Find the volume of the following solid regions. The solid above the region \(R=\\{(x, y): 0 \leq x \leq 1\), \(0 \leq y \leq 2-x\\}\) and between the planes \(-4 x-4 y+z=0\) and \(-2 x-y+z=8\)

Short answer

Answer: The volume of the solid region is \(\frac{2}{3}\) cubic units.

Step-by-step solution

Find the height function

To find the height of the region between the two given planes, subtract the z-values of the two planes. The given planes are: - \(z_1 = 4x + 4y\) - \(z_2 = 2x + y + 8\) Thus, the height function (h) is given by \(h = z_2 - z_1 = (2x + y + 8) - (4x + 4y) = -2x - 3y + 8\). Now, we will set up a double integral over the region R using the height function.

Set up the double integral over region R

The given region R is defined by \(0 \leq x \leq 1\) and \(0 \leq y \leq 2-x\). We'll set up the double integral of the height function over this region: \(\int_{0}^{1} \int_{0}^{2-x} (-2x - 3y +8) dy dx\)

Evaluate the integral with respect to y

To evaluate the integral with respect to y, perform the following: \(\int_{0}^{1} \left[-2x(y) - \frac{3}{2}(y^2) + 8y\right]_{0}^{2-x} dx\) First, let's plug in the limits for y: \(\int_{0}^{1} (-2x(2-x)-\frac{3}{2}(2-x)^2 +8(2-x)) dx\) Simplify the expression: \(\int_{0}^{1} (-4x + 2x^2 - 3(2-x)^2 + 16 - 8x) dx\)

Evaluate the integral with respect to x

Now we need to integrate with respect to x: \(\left[\int_{0}^{1} (2x^2 - 12x + 16 - 3(4-4x+x^2)) dx\right]\) Simplify the expression: \(\left[\int_{0}^{1} (2x^2 - 12x + 16 -12 +12x-3x^2) dx\right]\) Combine like terms: \(\left[\int_{0}^{1} (-x^2+1) dx\right]\) Integrate and evaluate the definite integral: \(\left[\frac{-x^3}{3} + x\right]_{0}^{1} = \frac{-1}{3} + 1 = \frac{2}{3}\) So, the volume of the solid region is \(\frac{2}{3}\) cubic units.

Key concepts

Double Integrals

Double integrals extend the concept of integration, typically used to find the area under a curve, into two dimensions. Instead of calculating area, they are used to compute volumes beneath surfaces over a certain region.

In a typical scenario, if we have a function that describes the height above each point

• The double integral sums up the values over the entire region.
• For a rectangular region, the setup of a double integral requires specifying limits for both variables involved.
• The order of integration (i.e., whether you integrate with respect to x or y first) can depend on the limits of the region.

In summary, comprehending double integrals helps in analyzing complex situations where we need to compute the volume under a surface over a specified area.

Solid Regions

Solid regions in calculus are three-dimensional sections of space that are bounded by surfaces or planes. In the problem context, the solid region is identified by its boundaries in the xy-plane and the height between two surfaces over this region.

When working with such regions, you define a region in the xy-plane, known as the base of the solid. Once defined:

• The region's shape can be any - rectangular, triangular, or any bounded shape.
• When finding volume, one essentially adds up small slabs over the base, each with a tiny area and height given by the distance between the bounding surfaces.

Understanding these setups is essential for visualizing how volumes are accumulated in real-world geometrical contexts.

Planes in Calculus

Planes in calculus represent flat two-dimensional surfaces within three-dimensional space. In calculus problems, these planes often define boundaries and constraints for regions of integration.

• A plane can be defined by a linear equation in three variables, typically written as Ax + By + Cz = D.
• In this exercise, the two planes given define upper and lower boundaries for the solid region.
• The intersection of planes with the base region (in the xy-plane) determines the limits of integration.

Knowing how planes interact in three dimensions is crucial for setting up and solving integrals to find enclosed volumes.

Calculus Integration

Calculus integration is a fundamental tool for solving problems related to areas, volumes, and other quantities involving rates of change and summation. In this exercise, integration is used to calculate the volume between two planes.

• Begin by breaking the volume problem into more manageable parts - calculating the integral in terms of one variable before the other.
• While setting up the integral, substitute the boundaries determined by the specific region involved.
• Compilation of the volume involves a step-by-step simplification and execution of the integral.

By integrating step by step, it becomes easier to understand the process, providing a clear path from problem setup to solution.