/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Evaluate the following integrals... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{0}^{2} \int_{1}^{2} \int_{0}^{1} y z e^{x} d x d z d y$$

Short Answer

Expert verified
Question: Evaluate the following triple integral:$$ \int_{0}^{1} \int_{1}^{2} \int_{0}^{2} yze^{x} dy \ dz \ dx $$ Answer: The evaluated triple integral is:$$ 6(e-1) $$

Step by step solution

01

Integrate with respect to \(x\)

Integrate the given function with respect to \(x\) using the limits of integration \(0\) and \(1\).$$ \int_{0}^{1} y z e^{x} d x $$Next, apply the integration formula:$$ e^x \int_{0}^{1} y z d x $$Now, evaluate the integral using the limits of integration from \(0\) to \(1\):$$ yze^1 - yze^0 = yz(e - 1) $$
02

Integrate with respect to \(z\)

Integrate the resulting function with respect to \(z\) using the limits of integration \(1\) and \(2\).$$ \int_{1}^{2} yz(e - 1) d z $$Apply the integration formula:$$ \frac{1}{2}y(e-1)(z^2) \bigg|_{1}^{2} $$Now, evaluate the integral using the limits of integration from \(1\) to \(2\):$$ [(\frac{1}{2}y(e-1)(2^2)) - (\frac{1}{2}y(e-1)(1^2))] $$
03

Integrate with respect to \(y\)

Integrate the resulting function with respect to \(y\) using the limits of integration \(0\) and \(2\).$$ \int_{0}^{2} [(\frac{1}{2}y(e-1)(2^2)) - (\frac{1}{2}y(e-1)(1^2))] d y $$Apply the integration formula:$$ \frac{1}{4}(e-1)((2^2)(y^2) - (y^2)) \bigg|_{0}^{2} $$Now, evaluate the integral using the limits of integration from \(0\) to \(2\):$$ \frac{1}{4}(e-1)((2^2)(2^2) - (2^2)) = \frac{3}{4}(e - 1)(8) $$
04

Final Answer

The final evaluated triple integral is:$$ \frac{3}{4}(e - 1)(8) = 6(e-1) $$

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Key Concepts

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

Evaluating Integrals
When dealing with integrals, it's important to understand that evaluation refers to the process of finding the numerical value of the integral. In this exercise, we were tasked with evaluating a triple integral, which involves calculating an integral in three distinct stages with each stage corresponding to one of the variables: \(x\), \(z\), and \(y\).

We started with the innermost integral, \(\int_{0}^{1} y z e^{x} d x\), integrating first with respect to \(x\). This required using the exponential integration rule that simplifies to \(e^{x}\) evaluated between 0 and 1. This calculation reduced our expression to \(yz(e - 1)\).

The next step was evaluating the integral with respect to \(z\). Again, we applied the limits given from 1 to 2. Finally, the outermost integral, with respect to \(y\), used similar techniques. As you move from the innermost to the outermost integrals, keep substituting back from the calculated expressions to always refine the integral and maintain integration bounds. The ultimate goal is to simplify the function entirely and arrive at a numerical value.
Iterated Integrals
When evaluating multiple integrals, especially in calculus, it is common to perform what is known as iterated integration. This involves evaluating each integral one at a time, starting with the innermost integral and moving systematically outward.

In our triple integral scenario, the process was to first evaluate the integral with respect to \(x\) while treating \(y\) and \(z\) as constants. Subsequently, we worked on the resulting expression for \(z\), treating \(y\) as constant. Lastly, \(y\) integration was executed which concluded the iterated integration process. This step-by-step approach helps manage complex problems and streamline the calculation process to support deep understanding of how each part contributes to the final answer.

Iterated integrals encapsulate a strategy of reducing multidimensional integration problems by breaking them down into stepwise single-variable problems, simplifying the computational challenge that continuous integrals can pose.
Integration Techniques
Understanding integration techniques is crucial when tackling problems like this. Knowing when and how to apply specific rules and formulas can make the task of integrating more manageable.

In this problem, the primary technique used is the substitution rule for the exponential function, where we integrated \(e^{x}\) between certain limits. It's vital to be familiar with common integrals, such as those of exponential, logarithmic, and polynomial functions, and how to use these to simplify expressions.

Another technique is recognizing that constants can be factored out of the integral to simplify the process, which we see in evaluating \(yz(e-1)\) while integrating with respect to \(x\). Further, careful selection of the order of integration can simplify the problem significantly, especially in cases of complex or non-standard regions of integration. Choosing wisely can result in simpler bounds or reduce complexity, as was exemplified in the conclusion of this triple integration task where the expression could be simplified to a straightforward arithmetic computation.

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Most popular questions from this chapter

The Jacobian is a magnification (or reduction) factor that relates the area of a small region near the point \((u, v)\) to the area of the image of that region near the point \((x, y)\) a. Suppose \(S\) is a rectangle in the \(u v\) -plane with vertices \(O(0,0)\) \(P(\Delta u, 0),(\Delta u, \Delta v),\) and \(Q(0, \Delta v)\) (see figure). The image of \(S\) under the transformation \(x=g(u, v), y=h(u, v)\) is a region \(R\) in the \(x y\) -plane. Let \(O^{\prime}, P^{\prime},\) and \(Q^{\prime}\) be the images of O, P, and \(Q,\) respectively, in the \(x y\) -plane, where \(O^{\prime}\) \(P^{\prime},\) and \(Q^{\prime}\) do not all lie on the same line. Explain why the coordinates of \(\boldsymbol{O}^{\prime}, \boldsymbol{P}^{\prime}\), and \(Q^{\prime}\) are \((g(0,0), h(0,0))\) \((g(\Delta u, 0), h(\Delta u, 0)),\) and \((g(0, \Delta v), h(0, \Delta v)),\) respectively. b. Use a Taylor series in both variables to show that $$\begin{aligned} &g(\Delta u, 0) \approx g(0,0)+g_{u}(0,0) \Delta u\\\ &g(0, \Delta v) \approx g(0,0)+g_{v}(0,0) \Delta v\\\ &\begin{array}{l} h(\Delta u, 0) \approx h(0,0)+h_{u}(0,0) \Delta u \\ h(0, \Delta v) \approx h(0,0)+h_{v}(0,0) \Delta v \end{array} \end{aligned}$$ where \(g_{u}(0,0)\) is \(\frac{\partial x}{\partial u}\) evaluated at \((0,0),\) with similar meanings for \(g_{v}, h_{u}\) and \(h_{v}\) c. Consider the vectors \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\) and the parallelogram, two of whose sides are \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\). Use the cross product to show that the area of the parallelogram is approximately \(|J(u, v)| \Delta u \Delta v\) d. Explain why the ratio of the area of \(R\) to the area of \(S\) is approximately \(|J(u, v)|\)

Consider the thin constant-density plate \(\\{(r, \theta): 0 < a \leq r \leq 1,0 \leq \theta \leq \pi\\}\) bounded by two semicircles and the \(x\) -uxis. a. Find and graph the \(y\) -coordinate of the center of mass of the plate as a function of \(a\) b. For what value of \(a\) is the center of mass on the edge of the plate?

Evaluate the Jacobians \(J(u, v, w)\) for the following transformations. $$x=v w, y=u w, z=u^{2}-v^{2}$$

A right circular cylinder with height \(8 \mathrm{cm}\) and radius \(2 \mathrm{cm}\) is filled with water. A heated filament running along its axis produces a variable density in the water given by \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \mathrm{g} / \mathrm{cm}^{3}(\rho\) stands for density here, not the radial spherical coordinate). Find the mass of the water in the cylinder. Neglect the volume of the filament.

Use a double integral to compute the area of the following regions. Make a sketch of the region. The region bounded by the parabola \(y=x^{2}\) and the line \(y=x+2\)

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