Problem 15 - Calculus Early Transcendentals

Question

Find the volume of the following solids using triple integrals. The solid in the first octant bounded by the plane \(2 x+3 y+6 z=12\) and the coordinate planes

Step-by-Step Solution

Answer

Question: Find the volume of the solid bounded by the plane \(2x+3y+6z=12\) and the coordinate planes in the first octant. Answer: The volume of the solid is 48 cubic units.

Step 1: Determine Limits of Integration for x, y, and z

In the first octant, all coordinates are non-negative: \(x\ge0\), \(y\ge0\), and \(z\ge0\). The coordinate planes are given by \(x=0\), \(y=0\), and \(z=0\). To find the limits of integration, we need to find the intersections between the plane \(2x+3y+6z=12\) and each coordinate plane. When \(y=0\) and \(z=0\), we have \(2x=12\). Thus, \(x=6\). When \(x=0\) and \(z=0\), we have \(3y=12\). Thus, \(y=4\). When \(x=0\) and \(y=0\), we have \(6z=12\). Thus, \(z=2\). Hence, the limits of integration for x, y, and z are \(0\le x\le 6\), \(0\le y\le 4\), and \(0\le z\le 2\).

Step 2: Integration over z

First, we integrate the function over z with the limits \(0\le z\le 2\). Since the volume is given by \(1 \cdot dV\), we have: \(\int_0^2 \int_0^4 \int_0^6 1\, dx\, dy\, dz\).

Step 3: Integration over x

The next step is to integrate over x with the limits \(0\le x\le 6\). We have: \(\int_0^2 \int_0^4 [\int_0^6 1\, dx] dy\, dz = \int_0^2 \int_0^4 [x]_{0}^6 dy\, dz = \int_0^2 \int_0^4 6\, dy\, dz\).

Step 4: Integration over y

Now, we need to integrate over y with the limits \(0\le y\le 4\). We have: \(\int_0^2 [\int_0^4 6\, dy] dz = \int_0^2 [6y]_{0}^4 dz = \int_0^2 24\, dz\).

Step 5: Integration over z to find the volume

Finally, we must integrate over z with the limits \(0\le z\le 2\). We have: \([\int_0^2 24\, dz] = [24z]_{0}^2 = 24(2) - 24(0) = 48\). Therefore, the volume of the solid is 48 cubic units.

Key Concepts

Limits of Integration

Understanding the limits of integration is crucial when calculating the volume of a region using triple integrals. The limits tell us the range over which we need to perform the integration to account for the entire volume of the solid. When defining these limits for a triple integral, we start by identifying the boundaries of the solid in each coordinate direction: x, y, and z.

For a solid in the first octant, the limits are non-negative since the first octant is where all three coordinates x, y, and z are positive. The coordinate planes act as the lower limits, which are always zero in this case. The upper limits are found by setting the remaining variables to zero, resulting in the intersection of the solid with the coordinate axes.

Using the given plane equation, we can find these upper limits by isolating each variable in turn, as seen in the exercise. The result is a set of inequalities: \(0 \leq x \leq 6\), \(0 \leq y \leq 4\), and \(0 \leq z \leq 2\), that describe the region over which we will integrate. This step is the foundation of solving the problem, as it delineates the volume within which we will carry out further integration.

First Octant Region

The first octant region is a term used to describe the portion of three-dimensional space where all three coordinates (x, y, z) are positive. The reference to 'octant' comes from the fact that in three-dimensional space, there are eight such regions or 'octants' created by the intersection of the xyz-coordinate planes.

In many problems involving triple integrals, including the exercise provided, you're asked to find volumes within the first octant. This implicitly sets the lower limits of integration to zero, as it's the starting point for all positive coordinates. Additionally, the three coordinate planes—xy, xz, and yz—bound the region, making it easier to visualize and set up the volume integral.

To help solidify this concept, picture a cube in the first octant whose corners touch the origin and each of the coordinate axes; this cube's volume is encapsulated by the points where the plane intersects the axes. Your task when solving these exercises is to accurately define these touchpoints to set up your triple integral correctly.

Integration Order

The order of integration in triple integrals refers to the sequence in which we integrate with respect to each variable. It's essential to determine the correct order to perform the integrations based on the limits of integration. Generally, we would want the integration to proceed from the innermost integral to the outermost, echoing the sequence of dx, dy, and dz.

In our example, the integration order starts with x, as it's the innermost integral, followed by y, and then z, which is the outermost integral. During this process, you treat the limits for x as constants when integrating with respect to y, and both x and y limits as constants when integrating with respect to z. This step-by-step approach, known as iterated integration, allows us to handle one variable at a time, simplifying the process and leading to the volume calculation.

The choice of integration order can greatly affect the simplicity of the calculations. One should always look for an order that makes the integration process more straightforward, either by resulting in nicer functions to integrate or by aligning with the symmetries of the region.