Problem 48 - Calculus Early Transcendentals

Question

Find the volume of the following solids. The solid beneath the plane \(f(x, y)=6-x-2 y\) and above the region \(R=\\{(x, y): 0 \leq x \leq 2,0 \leq y \leq 1\\}\)

Step-by-Step Solution

Answer

Answer: The volume of the solid is 6 cubic units.

Step 1: Set up the double integral

The double integral will be set up for function \(f(x, y) = 6 - x - 2y\) over the given rectangular region R with bounds for x and y \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\). The double integral will be in the form: $$ V=\int_0^1\int_0^2 (6-x-2y) \, dx \, dy $$

Step 2: Integrate with respect to x

We first integrate the inner integral with respect to x: $$ \int_0^2 (6-x-2y) \, dx = \Big[6x - \frac{1}{2}x^2 - 2xy\Big]_0^2 $$ Now, plug in the limits of x into the expression and simplify: $$ \Big[6(2) - \frac{1}{2}(2)^2 - 2(2)y\Big] - \Big[6(0) - \frac{1}{2}(0)^2 - 2(0)y\Big] = 12-4-4y $$ The result of the inner integral is \(12-4-4y\). Now, the double integral will look like: $$ V=\int_0^1 (12-4-4y) \, dy $$

Step 3: Integrate with respect to y

We now integrate the outer integral with respect to y: $$ \int_0^1 (12 - 4 - 4y) \, dy = \Big[12y - 4y - 2y^2\Big]_0^1 $$ Now, plug in the limits of y into the expression and simplify: $$ \Big[12(1) - 4(1) - 2(1)^2\Big] - \Big[12(0) - 4(0) - 2(0)^2\Big] = 12 - 4 - 2 $$

Step 4: Find the volume

After evaluating the double integral, we have the volume of the solid: $$ V=12-4-2=6 $$ The volume of the solid beneath the plane \(f(x, y) = 6 - x - 2y\) and above the region R is 6 cubic units.

Key Concepts

Double Integral

When calculating the volume of a solid, we often use double integrals. A double integral allows us to integrate over a two-dimensional area and is particularly used to compute volumes of solids bounded by surfaces. Here's how it works:

The double integral \( \int\int f(x, y) \,dx\,dy \) combines two single integrals into one. When you see \( \int \int \) it implies integration over a region in the xy-plane. The function \( f(x, y) \) represents the height of the solid above each point \( (x, y) \) in the region. The order of integration, which can be either \( \int dx \int dy \) or \( \int dy \int dx \) depending on the problem, indicates which variable is integrated first.

In our exercise, the solid is defined by the plane \( f(x, y) = 6 - x - 2y \) and is bounded by a rectangular region R. We calculate its volume by setting up the double integral over the region R and integrating the function that defines the height of the solid at each point within R.

• The inner integral is the integration with respect to x, meaning we find the area in the x-direction at each slice y.
• The outer integral takes the result of the inner integral and integrates with respect to y, effectively stacking these slices to find the total volume.

By applying the double integral to this problem, we were able to determine that the solid's volume is 6 cubic units.

Rectangular Region Integration

When dealing with rectangular region integration, the limits of integration are constants, simplifying the computation. The integrals can be easily solved by evaluating the integral from the lower limit to the upper limit for both x and y dimensions.

The given rectangular region R is described by the bounds \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \) in our exercise. Since these are constant limits, the region forms a rectangle on the xy-plane. This simplicity means that:

• There is no need to solve for variable bounds, as they are given outright.
• Integration can proceed straight away on each variable separately, which is known as iterated integration.
• The order of integration (whether you integrate with respect to x or y first) does not affect the final result due to the rectangular nature of the domain.

In practical terms, what this means is the integral for the volume can be set up as \( V=\int_a^b\int_c^d f(x, y) \,dx \,dy \) for our plane. That's exactly what we did: integrated first with x within \( [0, 2] \) and then with y within \( [0, 1] \) to find our result.

Triple Integration for Volume

From Double to Triple Integration

Though not directly applicable to our textbook exercise, triple integration is the three-dimensional counterpart to double integration. Just as a double integral can find the volume under a surface over a region in the xy-plane, a triple integral extends this concept to calculate the volume of a solid in three-dimensional space, accounting for x, y, and z dimensions.

The format for a triple integral is \( \int\int\int f(x, y, z) \,dx\,dy\,dz \) and it's used when the solid has variable bounds along the z-axis, not just confined to being between a function and the xy-plane. It's particularly useful for calculating volumes of more complex shapes, such as those bounded by curved surfaces.

• In triple integration, we first find the area in one variable, then the volume in two variables, and finally the volume in all three variables.
• The process of setting up a triple integral can involve changing the order of integration to make the calculations easier, similarly to double integrals, but with an added dimension of complexity.
• Applying triple integrals may require the use of cylindrical or spherical coordinates when the solid has symmetries that align with these coordinate systems.

Understanding triple integration elevates the concept of volume calculation to include a broad range of possible geometric formations, beyond the simple solids used in typical textbook exercises.