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 coordinate planes and \(z=2-2 x / 3-2 y\).

Short answer

The volume of region \(D\) is 1.

Step-by-step solution

Understand the Region

The region \(D\) is bounded by the coordinate planes \(x=0\), \(y=0\), \(z=0\), and the plane \(z = 2 - \frac{2x}{3} - 2y\). This is a tetrahedron with vertices at \((0,0,0)\), \((3,0,0)\), \((0,1,0)\), and \((0,0,2)\).

Set Limits for x First Integration Order

With \(z = 2 - \frac{2x}{3} - 2y\) as the top boundary, the limits for \(x\) are \(0\) to \(3\), for \(y\) they are \(0\) to \(\frac{3-3x}{6}\), and for \(z\) they are \(0\) to \(2 - \frac{2x}{3} - 2y\).

Set Limits for y First Integration Order

For the \(y\) first order, \(y\) goes from \(0\) to \(1\), \(x\) from \(0\) to \(3 - 3y\), and \(z\) from \(0\) to \(2 - \frac{2x}{3} - 2y\).

Set Limits for z First Integration Order

With \(z\) first, \(z\) ranges from \(0\) to \(2\), \(y\) from \(0\) to \(\frac{3 - 3z}{6}\), and \(x\) from \(0\) to \(3 - 3z - 6y\).

Triple Integrals for Each Order

The 6 triple integrals are: 1. \(\int_0^3 \int_0^{(3-3x)/6} \int_0^{2-\frac{2x}{3}-2y} dz \ dy \ dx \) 2. \(\int_0^3 \int_0^{2-\frac{2x}{3}} \int_0^{(3-3x)/6} dy \ dz \ dx \)3. \(\int_0^1 \int_0^{3-3y} \int_0^{2-\frac{2x}{3}-2y} dz \ dx \ dy \) 4. \(\int_0^1 \int_0^2 \int_0^{3-3y} dx \ dz \ dy \) 5. \(\int_0^2 \int_0^{(3-3z)/6} \int_0^{3-3z-6y} dx \ dy \ dz \)6. \(\int_0^2 \int_0^{3-3z} \int_0^{(3-3z-6y)/6} dy \ dx \ dz \).

Evaluate the First Integral

Choose the first integral for evaluation: \(\int_0^3 \int_0^{(3-3x)/6} \int_0^{2-\frac{2x}{3}-2y} dz \ dy \ dx \). Integrate with respect to \(z\), yielding \[(2 - \frac{2x}{3} - 2y)\] evaluated from \(z=0\), giving \(2 - \frac{2x}{3} - 2y\).

Integrate with respect to y

With the result from step 6 as the integrand, integrate with respect to \(y\) from \(0\) to \(\frac{3-3x}{6}\), this provides: \[\int_0^{(3-3x)/6} (2 - \frac{2x}{3} - 2y) dy = \left[(2y - \frac{2xy}{3} - y^2)\right]_0^{(3-3x)/6}\].

Integrate with respect to x

Substitute and simplify in step 7 to evaluate the final integral with respect to \(x\) from \(0\) to \(3\). Finally, integrate and solve, yielding \(1\) as the volume.

Key concepts

Volume of Tetrahedron

A tetrahedron is a three-dimensional shape with four triangular faces. Calculating the volume of such a shape can be done using triple integrals when the limits and the integration order are defined correctly. In this case, the given region is a triangular pyramid or tetrahedron bounded by coordinate planes and a slanted plane:

• The base is formed at the intersection of the plane with the xy-plane.
• The vertices are at \( (0,0,0), (3,0,0), (0,1,0), ext{ and } (0,0,2) \).

To find the volume, set up a triple integral. The integral encompasses the complete region within the bounds defined by the problem. Using integration, any function, such as the volume function for a region, can be evaluated by finding the limits of integration based on bounds that conform to the geometric properties of that particular shape.

Order of Integration

The order of integration is crucial when setting up and solving triple integrals. It determines in which order the integration over the variables \( x \), \( y \), and \( z \) will be performed. Specifically, for the volume of the tetrahedron, every variable has a limit depending on the other two. Changing the order of integration alters these limits significantly because the bounds relate to the geometry of the tetrahderon:

• The variables can be integrated in any order: \((x, y, z), (x, z, y), (y, x, z), (y, z, x), (z, x, y), ext{ or } (z, y, x)\).
• Different orders may sometimes simplify the evaluation, depending on function simplicity.

Each order will have unique integration limits, impacting the integration process. Thus, knowing how to switch orders confidently while setting the correct bounds ensures you calculate the integral accurately for any complex regions like a tetrahedron.

Limits of Integration

The limits of integration for the variables in a triple integral are crucial because they define the region of integration. In the case of the tetrahedron bounded by the coordinate planes and the plane \( z=2-2x/3-2y \), the limits need to adhere strictly to this geometry:

• The limits for \( x \) range depending on \( y ext{ and } z \)
• The limits for \( y \) adjust based on \( x ext{ and } z \)
• The limits for \( z \) are continually dependent on the other two variables, \( x ext{ and } y \)

For example, in one integration order, the bounds might be from \(x=0 ext{ to }3\) then \(y=0 ext{ to }\frac{3-3x}{6}\) and finally \(z=0 ext{ to }2-\frac{2x}{3}-2y\).Understanding and applying these limits can allow the region's volume to be correctly calculated without missing regions defined by the problem statement.

Coordinate Planes

Coordinate planes are fundamental when describing boundaries for three-dimensional shapes or regions like a tetrahedron. In this scenario, the coordinate planes are \( x=0 \), \( y=0 \), and \( z=0 \), which form part of the tetrahedron's interface. These planes act as limits for assistance in defining the tetrahedron:

• The \( x=0 ext{ and } y=0 \) planes are vertical planes that extend infinitely in other directions.
• The \( z=0 \) plane is the horizontal plane representing the xy-plane.

These boundaries create a finite region when used with the additional plane \( z=2-2x/3-2y \), which is slanted due to its equation. Defining these boundaries ensures that when calculating volumes or other properties, the integral stays within this confined region rather than unwanted regions. Identifying how coordinate planes intersect with other surfaces allows better setting up of any integration or geometric analysis.