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Chapter 5: Problem 194

In the following exercises, evaluate the triple integrals over the \(\quad\) bounded \(\quad\) region . $$ E=\left\\{(x, y, z) \mid a \leq x \leq b, h_{1}(x) \leq y \leq h_{2}(x), e \leq z \leq f\right\\} $$ \(\quad \varliminf_{E}(x y+y z+x z) d V\), where \(E=\left\\{(x, y, z) \mid 0 \leq x \leq 1,-x^{2} \leq y \leq x^{2}, 0 \leq z \leq 1\right\\}\)

Short Answer

Expert verified
The triple integral evaluates to \( \frac{2}{3} \).

Step by step solution

01

Set up the triple integral

The region \( E \) is defined by the inequalities \( 0 \leq x \leq 1 \), \( -x^2 \leq y \leq x^2 \), and \( 0 \leq z \leq 1 \). We need to evaluate the integral of \( (xy + yz + xz) \) over this region. The triple integral is set up as follows:\[ \int_{0}^{1} \int_{-x^2}^{x^2} \int_{0}^{1} (xy + yz + xz) \, dz \, dy \, dx \]
02

Integrate with respect to z

Integrate the function \( (xy + yz + xz) \) with respect to \( z \) from \( 0 \) to \( 1 \):\[ \int_{0}^{1} (xy + yz + xz) \, dz = x(yz + z^2) \Big|_{0}^{1} = (xy + y(1) + x(1)) - (0) = xy + y + x \]So, the expression becomes:\[ \int_{0}^{1} \int_{-x^2}^{x^2} (xy + y + x) \, dy \, dx \]
03

Integrate with respect to y

Now, integrate the expression \( xy + y + x \) with respect to \( y \) from \( -x^2 \) to \( x^2 \):\[ \int_{-x^2}^{x^2} (xy + y + x) \, dy = \left[ \frac{1}{2}xy^2 + \frac{1}{2}y^2 + xy \right]_{-x^2}^{x^2} \]Substituting the bounds:\[ \left( \frac{1}{2}x(x^2)^2 + \frac{1}{2}(x^2)^2 + x^2 \right) - \left( \frac{1}{2}x(-x^2)^2 + \frac{1}{2}(-x^2)^2 + (-x^2) \right) \]Simplifying:\[ \left( \frac{1}{2}x^5 + \frac{1}{2}x^4 + x^2 \right) - \left( \frac{1}{2}x^5 + \frac{1}{2}x^4 - x^2 \right) = 2x^2 \]Thus, it reduces to:\[ \int_{0}^{1} 2x^2 \, dx \]
04

Integrate with respect to x

Finally, integrate \( 2x^2 \) with respect to \( x \) from \( 0 \) to \( 1 \):\[ \int_{0}^{1} 2x^2 \, dx = \left[ \frac{2}{3}x^3 \right]_{0}^{1} = \frac{2}{3}(1)^3 - \frac{2}{3}(0)^3 = \frac{2}{3} \]
05

Conclusion: Final result

The value of the triple integral \( \int_{E}(xy + yz + xz) \, dV \) over the region \( E \) is \( \frac{2}{3} \).

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Key Concepts

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

Volume Integration
When you're dealing with volume integration, especially in the context of triple integrals, you're essentially calculating the volume of a three-dimensional region in space. For a specific example, consider evaluating integrals over a bounded region (in this case, the region \(E\)).

Volume integration involves stacking infinitesimally small pieces of volume on top of one another to fill up the entire space defined by your region. In our exercise, this is achieved by integrating the function \((xy + yz + xz)\) over \(x\), \(y\), and \(z\) axes within specified limits. This results in the computation of volume inside a three-dimensional object.

Think of the triple integral as a way of summing up tiny volume elements, \(dV\), over the entire region \(E\). In practical terms, this process helps in analyzing multi-layered physical problems where volume calculations are essential, such as in physics to determine mass from density distributions.
Calculus
Calculus provides the tools to handle continuous change and is foundational in understanding triple integrals. It allows us to compute areas and volumes, track curves, and describe the behavior of functions.

Within our exercise, the calculus doesn't stop with integration. It requires initially understanding the function we're integrating, \((xy + yz + xz)\), and how it behaves over different sections of the defined volume. This comprehension is vital as it shapes the setup of your integrals over such regions.

Triple integrals like this one may initially look complex, but breaking down the steps 鈥� as demonstrated in the exercise 鈥� helps illuminate how these integrals work. This process underscores how calculus bridges different areas of mathematics to solve real-world problems, enabling us to describe volumes and other complex structures accurately.
Multivariable Integration
Multivariable integration extends the principles of single-variable calculus into higher dimensions. Here, we're not only concerned with going from one point to another but addressing changes across a plane or in space.

This exercise exemplifies multivariable integration as it demands integration over three different variables \(x\), \(y\), and \(z\). Each has its range, contributing to constructing the volume \(E\). By working through these layers - beginning with \(z\), then \(y\), finally \(x\) 鈥� we're effectively collapsing this three-dimensional formulation into a tangible value.

Multivariable calculus is powerful. It informs fields like physics, engineering, and even economics where complex systems involve more than just a single factor. Thus, understanding multivariable integration is learning to view spaces comprehensively, noticing how individual elements influence and interact within a larger whole.

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Most popular questions from this chapter

Show that \(\iiint_{R} f\left(\sqrt{16 x^{2}+4 y^{2}+z^{2}}\right) d V=\frac{\pi}{2} \int_{0} f(\rho) \rho^{2} d \rho, \quad\) where \(f\) is a continuous function on [0,1] and \(R\) is the region bounded by the ellipsoid \(16 x^{2}+4 y^{2}+z^{2}=1\) .

[T] Use a CAS to evaluate the integral \(\iint_{E}\left(x^{2}+y^{2}\right) d V\) where \(E\) lies above the paraboloid \(z=x^{2}+y^{2}\) and below the plane \(z=3 y\).

The midpoint rule for the triple integral III \(f(x, y, z) d V\) over the rectangular solid box \(B\) is a generalization of the midpoint rule for double integrals. The region \(B\) is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum \(\sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f\left(\bar{x}_{i}, \bar{y}_{j}, \bar{z}_{k}\right) \Delta V,\) where \(\left(\bar{x}_{i}, \bar{y}_{j}, \bar{z}_{k}\right)\) is the center of the box \(B_{i j k}\) and \(\Delta V\) is the volume of each subbox. Apply the midpoint rule to approximate \(\varliminf_{B} x^{2} d V\) over the solid \(B=\\{(x, y, z)|0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1|\) by using a partition of eight cubes of equal size. Round your answer to three decimal places.

The following problems examine Mount Holly in the state of Michigan. Mount Holly is a landfill that was converted into a ski resort. The shape of Mount Holly can be approximated by a right circular cone of height \(1100 \mathrm{ft}\) and radius \(6000 \mathrm{ft}\). In reality, it is very likely that the trash at the bottom of Mount Holly has become more compacted with all the weight of the above trash. Consider a density function with respect to height: the density at the top of the mountain is still density \(400 \mathrm{lb} / \mathrm{ft}^{3}\) and the density increases. Every 100 feet deeper, the density doubles. What is the total weight of Mount Holly?

The following problems concern the Theorem of Pappus (see Moments and Centers of Mass (http://cnx.org/ content/m53649/latest/) for a refresher), a method for calculating volume using centroids. Assuming a region \(R\), when you revolve around the \(x\) -axis the volume is given by \(V_{x}=2 \pi A \bar{y},\) and when you revolve around the y-axis the volume is given by \(V_{y}=2 \pi A \bar{x},\) where \(A\) is the area of \(R\). Consider the region bounded by \(x^{2}+y^{2}=1\) and above \(y=x+1\). Find the volume when you revolve the region around the \(y\) -axis.
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