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Chapter 14: Problem 7

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{-2}^{2} \int_{3}^{6} \int_{0}^{2} d x d y d z$$

Short Answer

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Question: Evaluate the triple integral $\int_{-2}^{2}\int_{3}^{6}\int_{0}^{2} dxdydz$. Answer: 24

Step by step solution

01

Understand the region of integration

The given integral is: $$\int_{-2}^{2}\int_{3}^{6}\int_{0}^{2} dxdydz$$ This integral represents the volume of the region inside the rectangular prism formed by \(\{(x,y,z) | -2\leq x\leq2, 3\leq y\leq6,0\leq z\leq 2\}\). Since the integrand is 1, the value of the integral will be equal to the volume of the region.
02

Evaluate the triple integral

To evaluate this triple integral, we perform the integration one variable at a time, starting with the innermost integral (integrating with respect to \(x\)) then moving outwards (integrating with respect to \(y\) and finally, \(z\)). Let's evaluate the integral with respect to x first: $$\int_{0}^{2}1dx=[x]_{0}^{2}=2-0=2$$ Now we integrate the result with respect to y: $$\int_{3}^{6}2dy=[2y]_{3}^{6}=2(6-3)=6$$ Finally, integrate the result with respect to z: $$\int_{-2}^{2}6dz=[6z]_{-2}^{2}=6(2-(-2))=24$$ So the value of the triple integral is 24.

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Key Concepts

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

Triple Integrals
In calculus, a triple integral is a mathematical technique used to integrate a function of three variables over a three-dimensional region. Imagine it as summing up infinitely many tiny volume elements within a specified space. Each small volume element represents a small cube within the region.

The general form of a triple integral is \[ \int \int \int f(x, y, z) \, dx \, dy \, dz \]This expression means find the integral of function \( f(x, y, z) \) over the three-dimensional region defined by your limits. The final results can often represent a physical quantity such as mass, charge, or in many cases, volume.

The order of integration is crucial. In our problem, the integration is done first with respect to \( x \), then \( y \), and finally \( z \). Understanding the order helps us correctly interpret and perform the calculations.
Volume Calculation
The concept of volume calculation through triple integrals allows us to compute the volume of complicated shapes and spaces in three dimensions. In the specific problem we have, the limits of integration give us a prism defined by the coordinates:
  • \(-2 \leq x \leq 2\)
  • \(3 \leq y \leq 6\)
  • \(0 \leq z \leq 2\)
With functions more complex than 1, triple integrals help calculate varied properties, such as density or pressure distributions.

However, in this case, since our integrand is 1, the triple integral directly gives us the volume of the prism. Each integration step essentially stacks slices over the coordinate defined by the limits, forming the complete three-dimensional shape.
Rectangular Prism Integration
Integrating over a rectangular prism is a straightforward application of triple integrals due to the constant limits. A rectangular prism is also known as a cuboid, characterized by having each face at right angles.

In the given problem, each limit represents a face of the prism. Here, the integrals simulate filling the space of a box or room, one layer at a time from back to front, side to side, and bottom to top, calculating the volume step-by-step.

To visualize, start at the "bottom layer" of our space defined by \(0 \leq z \leq 2\). Integrate \(x\) across from \(-2\) to 2, then integrate \(y\) from 3 to 6, completing the base area. Finally, integrate the height \(z\). You effectively build a complete volume, layer by layer. This systematic approach assures accuracy in finding the total space within the geometric shape.

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Most popular questions from this chapter

Find the coordinates of the center of mass of the following solids with variable density. The interior of the prism formed by \(z=x, x=1, y=4,\) and the coordinate planes with \(\rho(x, y, z)=2+y\)

Choose the best coordinate system and find the volume of the following solid regions. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. Find the volume of material remaining in a hemisphere of radius 2 after a cylindrical hole of radius 1 is drilled through the center of the hemisphere perpendicular to its base.

Use the definition for the average value of a function over a region \(R\) (Section 1 ), \(\bar{f}=\frac{1}{\text { area of } R} \iint_{R} f(x, y) d A\). Find the average value of \(z=a^{2}-x^{2}-y^{2}\) over the region \(R=\left\\{(x, y): x^{2}+y^{2} \leq a^{2}\right\\},\) where \(a>0\)

A spherical cloud of electric charge has a known charge density \(Q(\rho),\) where \(\rho\) is the spherical coordinate. Find the total charge in the interior of the cloud in the following cases. a. \(Q(\rho)=\frac{2 \times 10^{-4}}{\rho^{4}}, 1 \leq \rho<\infty\) b. \(Q(\rho)=\left(2 \times 10^{-4}\right) e^{-0.01 \rho^{3}}, 0 \leq \rho<\infty\)

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid cone has a base with a radius of \(a\) and a height of \(h\). How far from the base is the center of mass?
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