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Chapter 5: Problem 191

In the following exercises, evaluate the triple integrals over the bounded region $$E=\left\\{(x, y, z) | a \leq x \leq b, h_{1}(x) \leq y \leq h_{2}(x), e \leq z \leq f\right\\}$$ \(\iiint_{B} (2 x+5 y+7 z) d V\) where \(E=\\{(x, y, z) | 0 \leq x \leq 1,0 \leq y \leq-x+1,1 \leq z \leq 2\\}\)

Short Answer

Expert verified
The value of the triple integral is \(\frac{81}{8}\).

Step by step solution

01

Identify the Integration Limits

We have the region \(E\) defined by \(0 \leq x \leq 1\), \(0 \leq y \leq -x+1\), and \(1 \leq z \leq 2\). These will be the limits for our triple integral.
02

Set Up the Triple Integral

Using the region limits, set up the triple integral as follows: \[ \int_{x=0}^{1} \int_{y=0}^{-x+1} \int_{z=1}^{2} (2x + 5y + 7z) \, dz \, dy \, dx \] This expresses the evaluation order from the innermost (\(z\)) to the outermost (\(x\)).
03

Evaluate the Innermost Integral

Evaluate the integral with respect to \(z\): \[ \int_{z=1}^{2} (2x + 5y + 7z) \, dz = \left[ 2xz + 5yz + \frac{7}{2}z^2 \right]_{z=1}^{2} \] Evaluating, we get:\[ (2x \times 2 + 5y \times 2 + \frac{7}{2} \times (2^2)) - (2x \times 1 + 5y \times 1 + \frac{7}{2} \times (1^2)) \] resulting in \[ (4x + 10y + 14) - (2x + 5y + \frac{7}{2}) = 2x + 5y + \frac{21}{2} \]
04

Evaluate the Middle Integral

Now evaluate with respect to \(y\):\[ \int_{y=0}^{-x+1} (2x + 5y + \frac{21}{2}) \, dy \]Perform the integration:\[ \left[ 2xy + \frac{5}{2}y^2 + \frac{21}{2}y \right]_{y=0}^{-x+1} \]Evaluate at bounds:\[ \left(2x(-x+1) + \frac{5}{2} (-x+1)^2 + \frac{21}{2}(-x+1)\right)- (0)= \]Simplify to \[ (2x - 2x^2 + \frac{5}{2} - 5x + \frac{5}{2}x^2 + \frac{21}{2} - \frac{21}{2}x) \] results in\[ -\frac{3}{2}x^2 - \frac{19}{2}x + 13 \]
05

Evaluate the Outermost Integral

Finally, evaluate with respect to \(x\): \[ \int_{x=0}^{1} \left(-\frac{3}{2}x^2 - \frac{19}{2}x + 13\right) \, dx \]Integrate:\[ \left[ -\frac{1}{2}x^3 - \frac{19}{4}x^2 + 13x \right]_{x=0}^{1} \]Evaluate the expression:\[ \left(-\frac{1}{2}(1)^3 - \frac{19}{4}(1)^2 + 13(1) \right) - \left(0\right) = -\frac{1}{2} - \frac{19}{4} + 13 \]Simplify to obtain the result:\[ \frac{81}{8} \]
06

Final Answer

The value of the triple integral over the region \(E\) is \(\frac{81}{8}\).

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Key Concepts

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

Integration Limits
When working with triple integrals, setting the integration limits is a crucial step as it defines the 3D region over which you're integrating. In our example, the region given is \[E=\{(x, y, z) | 0 \leq x \leq 1, 0 \leq y \leq -x+1, 1 \leq z \leq 2\}\]. This tells us boundaries for each variable:
  • \(x\) ranges from 0 to 1, which represents the horizontal limits.

  • \(y\) ranges from 0 to \(-x+1\) which tells us the vertical limits. Here, \(-x+1\) acts as a line which confines \(y\) within a trapezoid-like region.

  • \(z\) ranges from 1 to 2, denoting the depth or height limits in our space.
Determining these boundaries correctly will ensure the integration is over the right region, capturing all the points to evaluate integrals accurately.
Evaluating Integrals
Evaluating triple integrals involves computing the integral step-by-step across each variable set within the given boundaries. In this exercise, the integration proceeds from the innermost variable to the outermost:
  • First, solve for \(z\) by integrating \(\int_{z=1}^{2} (2x + 5y + 7z) \, dz\). This step simplifies the expression by focusing only on the changes in the \(z\) direction, treating \(x\) and \(y\) as constants.

  • Next up, the integral with respect to \(y\) is computed. Here, \(y\)'s limits go from \(0\) to \(-x+1\) and the expression from the previous step is integrated in terms of \(y\).

  • Finally, tackle \(x\) integration with limits from \(0\) to 1. This last step consolidates all previous computations, combining them into a final value that represents the overall volume or other unified measure of the bounded 3D region under the function \(2x + 5y + 7z\).
Evaluating step-by-step allows precision, ensuring each variable is accounted for in the multi-dimensional space.
Bounded Regions
Triple integrals are integral in evaluating regions in three-dimensional space that are bounded by specified limits. For example, in the given region \(E=\{(x, y, z) | 0 \leq x \leq 1, 0 \leq y \leq -x+1, 1 \leq z \leq 2\}\), the boundaries create a specific sub-volume in XYZ space. Bounded regions are made up of:
  • Lower and upper bounds in \(x, y, z\) directions that define the edges of the region.

  • Planar or nonlinear surfaces, represented as functions or inequalities.

  • Limits that ensure the triple integral covers the finite region of interest without extending into unwanted areas.
Understanding the concept of bounded regions enables you to precisely and accurately set up the problem, ensuring any computed integration genuinely reflects the area beneath your specified 3D curve or surface.

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Most popular questions from this chapter

Convert the integral $$ \int_{-4}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \int_{-\sqrt{16-x^{2}-y^{2}}}^{\sqrt{16-x^{2}-y^{2}}}\left(x^{2}+y^{2}+z^{2}\right) d z d x d y $$ into an integral in spherical coordinates.

\([\mathrm{T}]\) Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates $$ \int_{0}^{2 \pi} \int_{3 \pi / 4}^{\pi / 4} \int_{0}^{1} \rho^{2} \sin \varphi d \rho d \varphi d \theta $$ Find the volume \(V\) of the solid. Round your answer to three decimal places.

Let \(Q\) be the solid situated outside the sphere \(x^{2}+y^{2}+z^{2}=z\) and inside the upper hemisphere \(x^{2}+y^{2}+z^{2}=R^{2}, \quad\) where \(R>1 .\) If the density of the solid is \(\rho(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}},\) find \(R\) such that the mass of the solid is \(\frac{7 \pi}{2}\)

For the following problems, find the specified area or volume. The volume of the solid that lies between the paraboloid \(z=2 x^{2}+2 y^{2}\) and the plane \(z=8\) .

[T] The average density of a solid \(Q\) is defined as \(\rho_{\text {ave}}=\frac{1}{V(Q)} \iiint_{Q} \rho(x, y, z) d V=\frac{m}{V(Q)}, \quad\) where \(\quad V(Q)\) and \(m\) are the volume and the mass of \(Q,\) respectively. If the density of the unit ball centered at the origin is \(\rho(x, y, z)=e^{-x^{2}-y^{2}-z^{2}},\) use a CAS to find its average density. Round your answer to three decimal places.
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