Chapter 14: Problem 7
Compute the value of the triple integral $$ \iiint_{\Gamma} f(x, y, z) d V $$ \(f(x, y, z)=x y z: \quad T\) lies below the surface \(z=1-x^{2}\) and above the rectangle \(-1 \leqq x \leqq 1.0 \leqq y \leqq 2\) in the xy-plane.
Short Answer
Expert verified
The value of the triple integral is 0.
Step by step solution
01
Identify the Limits of Integration for z
The surface given is described by the equation \(z = 1 - x^2\). Since the region \(T\) is above the rectangle in the xy-plane and below this surface, the limits for \(z\) are from \(0\) to \(1 - x^2\).
02
Determine the Limits of Integration for y
The region in the xy-plane that is described as a rectangle gives the limits for \(y\). These are from \(0\) to \(2\), which is given directly in the problem description.
03
Establish the Limits of Integration for x
The limits for \(x\) are given as part of the rectangle in the problem statement. They are from \(-1\) to \(1\).
04
Set Up the Triple Integral
With the established limits, the integral becomes \[ \int_{-1}^{1} \int_{0}^{2} \int_{0}^{1-x^2} x y z \, dz \, dy \, dx \].
05
Integrate with Respect to z
Integrate \(x y z\) with respect to \(z\) from \(0\) to \(1 - x^2\):\[ \int_{0}^{1-x^2} x y z \, dz = \left[ \frac{1}{2} x y z^2 \right]_{0}^{1-x^2} = \frac{1}{2} x y (1-x^2)^2 \].
06
Integrate with Respect to y
Substitute the result from Step 5 into the next integral and integrate with respect to \(y\) from \(0\) to \(2\):\[ \int_{0}^{2} \frac{1}{2} x y (1-x^2)^2 \, dy = \frac{1}{2} x (1-x^2)^2 \left[ \frac{y^2}{2} \right]_{0}^{2} = \frac{1}{2} x (1-x^2)^2 \cdot 2^2 = 2 x (1-x^2)^2 \].
07
Integrate with Respect to x
Finally, integrate with respect to \(x\) from \(-1\) to \(1\):\[ \int_{-1}^{1} 2 x (1-x^2)^2 \, dx \].This integral can be expanded and simplified to \[ 2 \int_{-1}^{1} x (1 - 2x^2 + x^4) \, dx \].Evaluate each term separately:- \(2 \int_{-1}^{1} x \, dx = 0\)- \(-4 \int_{-1}^{1} x^3 \, dx = 0\)- \(2 \int_{-1}^{1} x^5 \, dx = 0\)The integral evaluates to 0, since the function is odd and limits are symmetric about zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits of Integration
In a triple integral, the limits of integration define the boundaries of the region over which you are integrating. When dealing with a region in three-dimensional space, it’s essential to break down this volume into manageable slices. Each of the three variables - in our case, x, y, and z - will have its own range (or limits) according to the problem's description.
- For z: The surface equation, \(z = 1 - x^2\), indicates that z ranges from 0 to \(1 - x^2\), creating a dome-like shape over the xy-plane rectangle.
- For y: From the given rectangle, y ranges from 0 to 2. This essentially adds length to our slice in the y-direction.
- For x: Lastly, x ranges from -1 to 1, defining the width of the slice, taken directly from the rectangle's boundaries on the xy-plane.
Iterated Integration
Iterated Integration means performing multiple integrals in sequence. In multivariable calculus, this technique is often used for handling integrals of functions with more than one variable. When calculating a triple integral, you will integrate in stages, handling one variable at a time, following the order given in the integral.
Consider starting with the innermost integral. By integrating \(f(x, y, z) = x y z\) with respect to z first, we reduce a three-dimensional problem into a two-variable situation. Then comes the integration with respect to y, and finally with x.
Consider starting with the innermost integral. By integrating \(f(x, y, z) = x y z\) with respect to z first, we reduce a three-dimensional problem into a two-variable situation. Then comes the integration with respect to y, and finally with x.
- The key is to take the integral of one variable fully, substituting the results into the next integration to simplify the computational complexity step by step.
- This process is also known as carrying out the integral in an 'iterated' or ordered manner.
Odd Function Symmetry
Odd function symmetry is a feature that simplifies certain integrals significantly. A function \(f(x)\) is odd if \(f(-x) = -f(x)\), meaning it has point symmetry about the origin on a graph.
In the context of definite integrals, especially those symmetric about zero, this property has an important implication: the integral of an odd function over a symmetric interval is zero. This is the principle applied in the final steps of the solution provided: \[2 \int_{-1}^{1} x(1 - 2x^2 + x^4) \, dx = 0\].
In the context of definite integrals, especially those symmetric about zero, this property has an important implication: the integral of an odd function over a symmetric interval is zero. This is the principle applied in the final steps of the solution provided: \[2 \int_{-1}^{1} x(1 - 2x^2 + x^4) \, dx = 0\].
- Evaluating each term separately, functions like \(x\), \(x^3\), and \(x^5\) are odd functions, simplifying our computation.
- Knowing when and how to use symmetry saves a considerable effort in computation by bypassing unnecessary calculations.
Multivariable Calculus
Multivariable calculus extends the concepts learned in single-variable calculus into multiple dimensions, allowing us to analyze functions involving several variables. One key area is understanding and computing triple integrals, which measure volumes and accumulated quantities across three-dimensional regions.
Some essential components include:
Some essential components include:
- Domains and Regions: These define the scope of integration, requiring understanding of geometric shapes and boundaries in \(\mathbb{R}^3\).
- Functions with Multiple Variables: Such functions, like \(f(x, y, z)\), incorporate compound relationships among variables, often requiring partial derivatives for deeper insights.
- Coordinate Systems: The flexibility to use various systems such as Cartesian, cylindrical, or spherical coordinates helps tailor the approach to the geometry of the problem.