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Chapter 15: Problem 15

Evaluate the integrals in Exercises \(7-20\). $$ \int_{0}^{1} \int_{0}^{2-x} \int_{0}^{2-x-y} d z d y d x $$

Short Answer

Expert verified
The value of the integral is \(\frac{7}{6}\).

Step by step solution

01

Understand the Problem

We have a triple integral over the region defined by the bounds in the order of integration: dz, dy, dx. We need to integrate with respect to z first, then y, and finally x.
02

Integrate with Respect to z

The innermost integral to evaluate is \(\int_{0}^{2-x-y} dz\). Since there is no z term in the integral, we simply multiply the length of the z interval by 1: \[ \int_{0}^{2-x-y} dz = (2-x-y) - 0 = 2-x-y. \]
03

Substitute and Integrate with Respect to y

Now substitute the result from Step 2 into the next integral:\[ \int_{0}^{2-x} (2-x-y) dy. \] Split it into two parts: \[ \int_{0}^{2-x} (2-x) dy - \int_{0}^{2-x} y dy. \]For the first integral, we have:\[ (2-x)(y) \bigg|_{0}^{2-x} = (2-x)(2-x) - (2-x)(0). \] Which simplifies to \[ (2-x)^2. \] For the second integral, use the rule for integrating y, \[ y^2/2 \bigg|_0^{2-x} = \frac{(2-x)^2}{2} - 0. \] Combine both parts:\[ (2-x)^2 - \frac{(2-x)^2}{2} = \frac{(2-x)^2}{2}. \]
04

Simplify and Integrate with Respect to x

We substitute back into the integral with respect to x:\[ \int_{0}^{1} \frac{(2-x)^2}{2} dx. \]Simplifying this to \[ \frac{1}{2} \int_{0}^{1} (4 - 4x + x^2) dx. \]Distribute the fraction:\[ \frac{1}{2} \left( \int_{0}^{1} 4 dx - \int_{0}^{1} 4x dx + \int_{0}^{1} x^2 dx \right). \]Evaluate each integral: \[ \int_{0}^{1} 4 dx = 4x \big|_0^1 = 4(1) - 0 = 4, \]\[ \int_{0}^{1} 4x dx = 4 \cdot \frac{x^2}{2} \big|_0^1 = 2x^2 \big|_0^1 = 2(1) - 0 = 2, \] \[ \int_{0}^{1} x^2 dx = \frac{x^3}{3} \big|_0^1 = \frac{1}{3} - 0 = \frac{1}{3}. \]Now substitute back:\[ \frac{1}{2} (4 - 2 + \frac{1}{3}) = \frac{1}{2} (2 + \frac{1}{3}) = \frac{1}{2} \cdot \frac{7}{3} = \frac{7}{6}. \]
05

Conclude the Integral

After simplifying the integrated values and summing them up, the computed volume is:\[ \frac{7}{6}. \]

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Key Concepts

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

Multiple Integration
Multiple integration is a technique used to integrate functions with more than one variable. In essence, it's the extension of single-variable integration to functions of multiple variables, such as functions in two (double integrals) or three (triple integrals) dimensions. When you deal with a triple integral, as in the original exercise, you are essentially taking the integral of a function in a three-dimensional space.
  • First, you integrate with respect to the innermost variable. This is often referred to as the \'dz\' step in a set-up like \( \int dz \int dy \int dx \).
  • Next, you move outward, integrating with respect to the next variable \'dy\'.
  • Lastly, you work with the outermost variable, for instance, \'dx\', as in our original step-by-step solution.
Every layer of integration builds upon the previous one as we solve systematically from the inside out. This approach allows the calculation of cumulative functions across multi-dimensional spaces, like calculating volumes or masses spread over a region.
Bounds of Integration
When solving a multiple integral problem, it's crucial to understand and correctly implement the bounds of integration. These boundaries tell us where the integration should start and stop in a given direction and are often functions themselves when dealing with non-rectangular regions.
In the given exercise, each integral had specific limits:
  • The innermost integral \( \int_{0}^{2-x-y} dz \) showed that the integration with respect to \'z\' happens from \'z=0\' to some function \'z=2-x-y\'.
  • The next integral \( \int_{0}^{2-x} dy \) explains that \'y\' integration happens between \'y=0\' and \'y=2-x\'.
  • Finally, the outermost integral \( \int_{0}^{1} dx \) means that \'x\' integration takes place from \'x=0\' to \'x=1\'.
The integration limits define the shape of the region under consideration. Getting these bounds right is key to computing the integral over the correct part of space, ensuring accurate final results.
Volume Calculation
One of the powerful applications of triple integrals is in calculating volumes of regions in three dimensions. In this particular problem, you investigated a region defined by certain bounds and calculated the volume within.The final result was \( \frac{7}{6} \), which represents the 'size' or 'amount of space' occupied by the region. Here's a simplified breakdown of how we arrived at this:
  • First, the \'z\' integration simplified the innermost expression within the given bounds, providing a result that defined a cross-sectional area.
  • Then, integrating this function with respect to \'y\' gave the area of a region in 2D space.
  • Finally, we integrated the result with respect to \'x\', sweeping that area across the final dimension to achieve the volume measure.
Each step narrows down the problem from a generalized expression into a single numerical value, representing the total volume enclosed by the defined region. Understanding each dimension\'s integration individually and in sequence is essential to getting the right volume calculation.

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Most popular questions from this chapter

Use a CAS double-integral evaluator to estimate the values of the integrals in Exercises \(67-70 .\) $$ \int_{0}^{1} \int_{0}^{1} e^{-\left(x^{2}+y^{2}\right)} d y d x $$

a. Suppose that a solid right circular cone \(C\) of base radius \(a\) and altitude \(h\) is constructed on the circular base of a solid hemisphere \(S\) of radius \(a\) so that the union of the two solids resembles an ice cream cone. The centroid of a solid cone lies one-fourth of the way from the base toward the vertex. The centroid of a solid hemisphere lies three-eighths of the way from the base to the top. What relation must hold between \(h\) and \(a\) to place the centroid of \(C \cup S\) in the common base of the two solids?

The region in the first octant bounded by the coordinate planes, the plane \(x+y=4,\) and the cylinder \(y^{2}+4 z^{2}=16\) (GRAPH NOT COPY)

(a) find the spherical coordinate limits for the integral that calculates the volume of the given solid and (b) then evaluate the integral. The solid bounded below by the hemisphere \(\rho=1, z \geq 0,\) and above by the cardioid of revolution \(\rho=1+\cos \phi\)

Let \(D_{\mathrm{u}} f\) denote the derivative of \(f(x, y)=\left(x^{2}+y^{2}\right) / 2\) in the direction of the unit vector \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j}\) a. Finding average value Find the average value of \(D_{\mathrm{u}} f\) over the triangular region cut from the first quadrant by the line \(x+y=1 .\) b. Average value and centroid Show in general that the average value of \(D_{\mathbf{u}} f\) over a region in the \(x y\) -plane is the value of \(D_{\mathbf{u}} f\) at the centroid of the region.
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