Question

Set up the triple integrals that give the volume of \(D\) in all 6 orders of integration, and find the volume of \(D\) by evaluating the indicated triple integral. \(D\) is bounded by the planes \(y=0, y=2, x=1, z=0\) and \(z=(3-x) / 2\) Evaluate the triple integral with order \(d x d y d z\).

Short answer

The volume of the region \( D \) is \( 2.5 \).

Step-by-step solution

Understanding the Boundaries

The region \( D \) is bounded by the planes and surfaces: \( y = 0 \), \( y = 2 \), \( x = 1 \), \( z = 0 \), and \( z = \frac{3-x}{2} \). These boundaries define a region in 3D space that we need to integrate over to find the volume.

Determine Limits of Integration for dx dy dz

For the order \( dx \, dy \, dz \), we first consider \( z \) varying from 0 to \( \frac{3-x}{2} \), then \( y \) varies from 0 to 2, and finally, \( x \) varies from 0 to 1. Thus, the triple integral is:\[\int_{0}^{1} \int_{0}^{2} \int_{0}^{\frac{3-x}{2}} dz \, dy \, dx\]

Evaluate the Integral over z

Integrate with respect to \( z \):\[\int_{0}^{\frac{3-x}{2}} 1 \, dz = \left[ z \right]_{0}^{\frac{3-x}{2}} = \frac{3-x}{2}\]

Evaluate the Integral over y

Substitute the result from Step 3 into the next integral and integrate with respect to \( y \):\[\int_{0}^{2} \frac{3-x}{2} \, dy = \left[ \frac{3-x}{2} \, y \right]_{0}^{2} = (3-x)\]

Evaluate the Integral over x

Substitute the result from Step 4 into the next integral and integrate with respect to \( x \):\[\int_{0}^{1} (3-x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{1} = \left(3 \times 1 - \frac{1^2}{2}\right) - 0 = 3 - \frac{1}{2} = 2.5\]

Conclusion: Volume of Region D

The volume of the region \( D \) is \( 2.5 \). The boundaries and the sequential integration over the variables \( z \), \( y \), and \( x \) sum up the total volume contained within the region.

Key concepts

Volume Calculation

Volume calculation using triple integrals is a powerful technique in multivariable calculus. It allows us to find the volume of three-dimensional shapes, even if their boundaries are irregular or defined by surfaces, rather than simple geometric shapes like cubes or spheres. By decomposing a volume into infinitely small parts, we can sum them up to get the total volume of a region. This is achieved through triple integration, where the integral is calculated over three dimensions.
For region \( D \) given by the boundaries \( y=0, y=2, x=1, z=0 \), and \( z=\frac{3-x}{2} \), integrating over these provides the total volume encapsulated within these bounds. Evaluating the integral correctly by following the established limits gives us the exact volume of the region.

Order of Integration

The order of integration in triple integrals refers to the sequence in which we integrate over different variables. For a Cartesian coordinate system, this often involves choosing between \( dx, dy, \) and \( dz \). There are six possible orders, but it’s important to choose one that simplifies computation.
For this exercise, we selected \( dx \, dy \, dz \) as the order of integration. This choice depended on the order in which the bounds of the region \( D \) are most naturally expressed, making the evaluation straightforward. Incorrect order selection can make computation difficult or even impossible, so starting with the right order helps streamline the entire process.

Limits of Integration

Limits of integration define the boundaries over which you are performing the integration. For the triple integral over region \( D \), these limits need to be carefully assigned to each variable.

• For \( z \), the limits are from 0 to \( \frac{3-x}{2} \), capturing the space between the plane \( z=0 \) and the surface \( z=(3-x)/2 \).
• For \( y \), the bounds are 0 to 2, coming from the planes \( y=0 \) and \( y=2 \).
• Finally, for \( x \), due to the vertical walls that are described in the problem, the limits are from 0 to 1.

Assigning these limits properly is crucial as it ensures the integration is performed over the correct region in 3D space.

Integration Techniques

Integration techniques used in triple integrals revolve around reducing complexity by evaluating one integral at a time. By integrating stepwise according to the chosen order \( dx \, dy \, dz \), each integral narrows down the space progressively until only a scalar remains.
In the given solution, integration began with \( z \), a simple bound difference, reducing the triple integral to a double integral.
Next, by integrating with respect to \( y \), the focus is confined to a plane, simplifying the calculation further.
Finally, the integral over \( x \) is evaluated, using basic techniques derived for single-variable integrals. By systematically tackling one variable at a time, complexity is minimized and precision is maximized.