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Chapter 11: Problem 7

Find the volume of the solid bounded by the planes \(\mathrm{x}=0, \mathrm{y}=0, \mathrm{z}=0\), and \(x+y+z=6\)

Short Answer

Expert verified
The volume of the solid bounded by the planes x=0, y=0, z=0, and x+y+z=6 can be found by evaluating the triple integral \( \int_{0}^{6}\int_{0}^{6-x}\int_{0}^{6-x-y} 1 dz dy dx \). The resulting volume is 108 cubic units.

Step by step solution

01

Write down the integration limits for x, y, and z

The integration limits for each coordinate are determined by the corresponding plane equations: - For x, we have x=0 and x+y+z=6 - For y, we have y=0 and y+z=6-x - For z, we have z=0 and z=6-x-y Thus, the limits of integration are: - x: 0 to 6 - y: 0 to 6-x - z: 0 to 6-x-y
02

Set up and solve the volume integral

We can find the volume using the triple integral of dV = dz dy dx. The integral will have the form: \( \displaystyle V = \int_{0}^{6}\int_{0}^{6-x}\int_{0}^{6-x-y} 1 dz dy dx \) Let us first solve the integral with respect to z: \( \displaystyle V = \int_{0}^{6}\int_{0}^{6-x} [(6-x-y) - 0] dy dx \) Now, we have a double integral: \( \displaystyle V = \int_{0}^{6}\int_{0}^{6-x} (6-x-y) dy dx \) Now, we integrate with respect to y: \( \displaystyle V = \int_{0}^{6}\left[ 6y - \frac{y^2}{2} - xy \right]_0^{6-x} dx \) Now, we plug in the limits and simplify the expression: \( \displaystyle V = \int_{0}^{6} (36 - 12x + x^2 - 18x + 3x^2) dx \) And combine the terms: \( \displaystyle V = \int_{0}^{6} (36 - 30x + 4x^2) dx \) Finally, we integrate with respect to x: \( \displaystyle V = \left[ 36x - 15x^2 + \frac{4}{3}x^3 \right]_0^6 \) Find the volume V by plugging the upper limit (6) into the equation: \( \displaystyle V = 36(6) - 15(6)^2 + \frac{4}{3}(6)^3 \) Calculate the result: \( \displaystyle V = 216 - 540 + 288 \) \( \displaystyle V = 108 \) The volume of the solid bounded by the planes x=0, y=0, z=0, and x+y+z=6 is 108 cubic units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triple Integral
The triple integral is an essential mathematical tool used to calculate the volume of three-dimensional spaces. Much like how a single integral represents area under a curve, a triple integral extends this concept to assess the magnitude within a three-dimensional region. In the exercise given, the function inside the triple integral is simply '1', because we are calculating volume and each point within the domain contributes equally to the total volume.

To evaluate a triple integral, one often integrates in stages, starting from the innermost integral and working outward. For instance, the provided solution begins by integrating with respect to the variable 'z', then 'y', and finally 'x'. Each integration step collapses one dimension, reducing the triple integral to a double integral, and then to a single integral, until a numerical volume is obtained. Understanding the order of integration and the limits for each variable is crucial to correctly solve a triple integral.
Integration Limits
Integration limits define the boundaries within which you'll integrate to find the volume of a solid. Setting up the correct integration limits is pivotal because these limits describe the exact region over which you're calculating the volume. In the context of triple integrals, there are three sets of limits, one for each variable ('x', 'y', and 'z').

In our example, the given planes such as x=0 (the yz-plane), y=0 (the xz-plane), z=0 (the xy-plane), and the plane x+y+z=6 define a closed bounded region in the first octant of a three-dimensional space. The limits for 'x' are from 0 to 6, for 'y' from 0 to 6-x, and for 'z' from 0 to 6-x-y. Each set of limits is dependent on the preceding variables, leading to a nested or iterated integral. It's this interdependence of the limits that students often find tricky, but grasping this concept is vital for finding the volume of complex solids.
Volume Calculation
The volume calculation using triple integrals boils down to integrating a function over a three-dimensional region. When the function is simply '1', as is the case with the provided exercise, we are finding the volume of the shape enclosed by the specified limits. The ultimate goal is to accumulate the contributions of infinitesimally small volume elements represented by 'dV = dz dy dx', over the entire region of interest.

Once the appropriate integration limits are established, as they were in the exercise with the limits for x, y, and z, the volume is found by successively performing the integrations with respect to each variable. The example shows the practical execution: after each integration, the integral becomes simpler, transitioning from three dimensions to one. Finally, by evaluating the resulting expression at the specified limits, one can determine the precise volume of the solid in question, which in our exercise is a neat 108 cubic units. Understanding this process is useful not only for academic exercises but also for real-world applications where space and volume calculations are necessary.

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Most popular questions from this chapter

(a) Graph \(r=1 /(4 \cos \theta)\) for \(-\pi / 2 \leq \theta \leq \pi / 2\) and \(r=1\). Then write an iterated integral in polar coordinates representing the area inside the curve \(r=1\) and to the right of \(r=1 /(4 \cos \theta)\). (Use \(t\) for \(\theta\) in your work.)

Evaluate the double integral \(I=\iint_{\mathbf{D}} x y d A\) where \(\mathbf{D}\) is the triangular region with vertices (0,0),(6,0),(0,2) .

(a) You are given the point \((1, \pi / 2)\) in polar coordinates. (i) Find another pair of polar coordinates for this point such that \(r>0\) and \(2 \pi \leq \theta<4 \pi\) \(r=\) \(\theta=\) (ii) Find another pair of polar coordinates for this point such that \(r<0\) and \(0 \leq \theta<2 \pi\) \(r=\) \(\theta=\) (b) You are given the point \((-2, \pi / 4)\) in polar coordinates. (i) Find another pair of polar coordinates for this point such that \(r>0\) and \(2 \pi \leq \theta<4 \pi\) \(r=\) \(\theta=\) (ii) Find another pair of polar coordinates for this point such that \(r<0\) and \(-2 \pi \leq \theta<0\) \(r=\) \(\theta=\) (c) You are given the point (3,2) in polar coordinates. (i) Find another pair of polar coordinates for this point such that \(r>0\) and \(2 \pi \leq \theta<4 \pi\) \(r=\) \(\theta=\) (ii) Find another pair of polar coordinates for this point such that \(r<0\) and \(0 \leq \theta<2 \pi\) \(r=\) $$ \theta= $$

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles \(x^{2}+y^{2}=36\) and \(x^{2}-6 x+y^{2}=0\)

Values of \(f(x, y)\) are given in the table below. Let \(R\) be the rectangle \(1 \leq x \leq 1.6,2 \leq y \leq 3.2\). Find a Riemann sum which is a reasonable estimate for \(\int_{R} f(x, y) d a\) with \(\Delta x=0.2\) and \(\Delta y=0.4 .\) Note that the values given in the table correspond to midpoints. $$ \begin{array}{|c||c|c|c|} \hline y \backslash x & 1.1 & 1.3 & 1.5 \\ \hline \hline 2.2 & 4 & 2 & -1 \\ \hline 2.6 & 5 & 2 & 0 \\ \hline 3.0 & 9 & -5 & 0 \\ \hline \end{array} $$ \(\int_{R} f(x, y) d a \approx\) _______________________.
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