Chapter 13: Problem 9
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
Expert verified
The volume of region \(D\) is 1.
Step by step solution
01
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 \((0,0,2)\).
02
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 \(2 - \frac{2x}{3} - 2y\).
03
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 \(2 - \frac{2x}{3} - 2y\).
04
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 \(3 - 3z - 6y\).
05
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. \(\int_0^2 \int_0^{3-3z} \int_0^{(3-3z-6y)/6} dy \ dx \ dz \).
06
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 \(2 - \frac{2x}{3} - 2y\).
07
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: \[\int_0^{(3-3x)/6} (2 - \frac{2x}{3} - 2y) dy = \left[(2y - \frac{2xy}{3} - y^2)\right]_0^{(3-3x)/6}\].
08
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( (0,0,0), (3,0,0), (0,1,0), ext{ and } (0,0,2) \).
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.
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, \( x ext{ and } y \)
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.