Problem 99 - Calculus Early Transcendentals

Question

Compute the volume of the following solids. The column with a square base \(R=\\{(x, y):|x| \leq 1,|y| \leq 1\\}\) cut by the plane \(z=4-x-y\)

Step-by-Step Solution

Answer

Question: Compute the volume of a solid with a square base and cut by a plane. The top plane is given by z = 4-x-y, and the base plane is z = 0. The square base has vertices (-1, 1), (1, 1), (1, -1), and (-1, -1). Answer: The volume of the given solid is 16 cubic units.

Step 1: Set up the Integral

We need to set up a double integral to find the volume. The integral will be over the region R with bounds from -1 to 1 in both the x and y directions. The height will be the difference between the top and bottom planes, which is \(4-x-y\). $$ V = \iint_R (4-x-y) \, dx\,dy $$

Step 2: Integrate with respect to x

Next, we will integrate the function with respect to x. The limits of integration for x will be from -1 to 1. $$ V = \int_{-1}^{1} \int_{-1}^{1} (4-x-y) \, dx\,dy $$ First, find the antiderivative with respect to x: $$ \int (4-x-y) \, dx = 4x - \frac{1}{2}x^2 - xy + C_y $$ Now, find the volume by evaluating the integral's limits: $$ V = \left[ 4x - \frac{1}{2}x^2 - xy \right]_{-1}^{1} = (4(1) - \frac{1}{2}(1)^2 - y(1)) - (4(-1) - \frac{1}{2}(-1)^2 + y(-1)) $$ Simplify the expression: $$ V = (4 - \frac{1}{2} - y) - (-4 - \frac{1}{2} + y) $$ $$ V = 8 - 2y $$ Keep this result, and move on to the next integral.

Step 3: Integrate with respect to y

Now, we will integrate the simplified expression with respect to y. The limits of integration for y will be from -1 to 1. $$ V = \int_{-1}^{1} (8 - 2y) \, dy $$ First, find the antiderivative with respect to y: $$ \int (8 - 2y) \, dy = 8y - y^2 + C $$ Now, find the volume by evaluating the integral's limits: $$ V = \left[ 8y - y^2 \right]_{-1}^{1} = (8(1) - (1)^2) - (8(-1) - (-1)^2) $$ Simplify the expression: $$ V = (8 - 1) - (-8 - 1) $$ $$ V = 16 $$ Hence, the volume of the given solid is 16 cubic units.

Key Concepts

Double Integral

The concept of the double integral is crucial when you want to calculate the volume of a solid with variable height over a particular region. In essence, a double integral allows you to account for variations across two dimensions—often denoted as 'x' and 'y'. To perform a double integral, you first integrate with respect to 'x', treating 'y' as a constant (or vice versa), and then integrate the resulting expression with respect to 'y'.

Imagine you're painting a textured wall where the amount of paint needed varies across the wall's surface. Just as you'd assess the paint needed for every section and sum it all up, the double integral sums up infinitesimally small areas ('dx' times 'dy') multiplied by the function that provides the height at each point, which gives you the total volume.

To ensure that students can follow along, it's important to clarify that each integration step considers the effect of one variable at a time, simplifying the process. By integrating with respect to 'x', we focus on the variable 'x' first, holding 'y' steady. Subsequently, integrating with respect to 'y' involves using the results from the first integration and considering the effect of ‘y’. The double integral represents the accumulation of all these tiny volumes to achieve the volume of the entire solid.

Volume by Integration

To find the volume of a solid using integration, visualize dividing the solid into infinitesimally thin slices, similar to slicing bread. Each slice has a certain area that can be found using an integral. By summing the volumes of these slices—effectively stacking them back together—you get the volume of the whole solid.

This approach is particularly useful when dealing with irregular shapes, where traditional geometric formulas for volume are inapplicable. For example, when calculating the volume of the given solid with a square base and cut by a plane, we cannot simply use the formula for a cube or prism. Instead, we integrate the area of each slice along the height of the solid.

Using integration, we transform a complex three-dimensional problem into a more manageable two-dimensional one, integrating over the base area and then considering the height of the solid at each point. This method provides a powerful way to tackle challenging volume problems in calculus.

Antiderivative

An antiderivative, sometimes called an indefinite integral, is essentially the reverse process of differentiation. If differentiation gives us the rate at which a function changes, then an antiderivative gives us the original function before it was differentiated. In finding volumes by integration, antiderivatives are essential because they allow us to revert our function back to an expression from which we can calculate areas or volumes.

For students, understanding antiderivatives means recognizing them as the 'accumulated sum' of the rates of change. Finding an antiderivative involves determining the function whose derivative would result in the given expression. This function typically has an added constant, often denoted as 'C', since the differentiation of a constant is zero.

In our volume calculation problem, we took the antiderivative of our integrand with respect to 'x' first and then with respect to 'y'. This step-by-step method simplifies the process by systematically tackling one dimension at a time, emphasizing the integral aspect of calculus, which is about accumulating change to find quantities like area, volume, and other summations.