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Chapter 5: Problem 208

In the following exercises, evaluate the triple integrals over the bounded region \(E=\left\\{(x, y, z) |(x, y) \in D, u_{1}(x, y) x \leq z \leq u_{2}(x, y)\right\\}\) where \(D\) is the projection of \(E\) onto the \(x y\) -plane. \(\iint_{D}\left(\int_{1}^{3} x(z+1) d z\right) d A where \) \(D=\left\\{(x, y) | x^{2}+y^{2} \leq 1\right\\}\)

Short Answer

Expert verified
The triple integral evaluates to 0.

Step by step solution

01

Identify the Limits of Integration for z

The problem specifies the inner integral's limits as 1 to 3, meaning that for any point \((x, y)\), the variable \(z\) ranges from 1 to 3. Therefore, for the triple integral, the limits for \(z\) are \(1 \leq z \leq 3\).
02

Setup the Inner Integral with respect to z

The expression for the inner integral is given by \(\int_{1}^{3} x(z+1) \, dz\). This integral needs to be evaluated next.
03

Perform the Inner Integral

Evaluate \(\int_{1}^{3} x(z+1) \, dz\):\[\int_{1}^{3} x(z+1) \, dz = x\int_{1}^{3} (z+1) \, dz = x \left[ \frac{z^2}{2} + z \right]_{1}^{3}.\]Calculate:\[= x \left[ \left( \frac{3^2}{2} + 3 \right) - \left( \frac{1^2}{2} + 1 \right) \right] = x \left[ \left( \frac{9}{2} + 3 \right) - \left( \frac{1}{2} + 1 \right) \right].\]
04

Simplify the Result of Inner Integral

Continuing from the previous step:\[x \left[ \left( \frac{9}{2} + 3 \right) - \left( \frac{1}{2} + 1 \right) \right] = x \left( \frac{9}{2} + \frac{6}{2} - \frac{1}{2} - \frac{2}{2} \right)= x \left( \frac{12}{2} \right)= 5x.\]
05

Describe Region D in the xy-plane

The region \(D\) is defined as \(x^2 + y^2 \leq 1\), which is a circle of radius 1 centered at the origin in the \(xy\)-plane.
06

Set Up the Double Integral over D

The next step is to evaluate the double integral over \(D\). Change the integral to polar coordinates: \(x = r \cos \theta\), \(y = r \sin \theta\), \(dA = r \, dr \, d\theta\). The region \(x^2 + y^2 \leq 1\) translates to \(0 \leq r \leq 1\), \(0 \leq \theta \leq 2\pi\).
07

Perform the Double Integral in Polar Coordinates

Transform the expression into polar coordinate form:\[\iint_D 5x \, dA = \int_{0}^{2\pi} \int_{0}^{1} 5(r \cos\theta) \cdot r \, dr \, d\theta.= \int_{0}^{2\pi} \int_{0}^{1} 5r^2 \cos\theta \, dr \, d\theta.\]
08

Solve the Inner Integral with respect to r

Evaluate the inner integral:\[\int_{0}^{1} 5r^2 \cos\theta \, dr = 5\cos\theta \int_{0}^{1} r^2 \, dr = 5\cos\theta \left[ \frac{r^3}{3} \right]_{0}^{1} = 5\cos\theta \cdot \frac{1}{3}= \frac{5}{3}\cos\theta.\]
09

Solve the Outer Integral with respect to 胃

Evaluate the outer integral:\[\int_{0}^{2\pi} \frac{5}{3}\cos\theta \, d\theta.\]The integral of \(\cos\theta\) over one full cycle \(0\) to \(2\pi\) is zero:\[\int_{0}^{2\pi} \cos\theta \, d\theta = 0.\]Thus, the entire integral evaluates to zero.

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Key Concepts

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

Projection in the XY-plane
When dealing with triple integrals, it's essential to understand the region over which you're integrating. The projection in the xy-plane is the shadow or outline of a 3D region onto the xy-plane.
This projection helps us determine the limits of integration for a double integral, which simplifies the 3D problem into 2D form.
  • In our problem, the projection is denoted as region \(D\).
  • Region \(D\) is where all points \((x, y)\) reside such that the condition \(x^2 + y^2 \leq 1\) holds true.
Understanding this projection helps set up the integral by defining a constrained, manageable area to evaluate.
Circle in Polar Coordinates
The region \(D\) in our exercise forms a circle in the xy-plane with radius 1 and centered at the origin. Describing this circle in polar coordinates makes the integration process much more intuitive and straightforward.
  • In polar coordinates, any point \((x, y)\) is expressed in terms of \(r\), the distance from the origin, and \(\theta\), the angle with the positive x-axis.
  • The transformation equations are \(x = r \cos \theta\) and \(y = r \sin \theta\).
  • The inequality \(x^2 + y^2 \leq 1\) translates to \(0 \leq r \leq 1\) and \(0 \leq \theta \leq 2\pi\).
This makes describing complex boundaries like circles much easier, facilitating further steps in integration.
Integration limits
Integration limits define the bounds of the integrals in the problem. They are crucial because they tell us the range over which we will sum the infinitesimal parts of the function.
First, we consider the limits for the variable \(z\):
  • The exercise specifies these as 1 to 3 for any \((x, y)\) point, indicating that \(z\) remains within this range throughout the entire region.
Next, the limits for \((x, y)\) as described by region \(D\):
  • Once transformed to polar coordinates, \(r\) varies from 0 to 1, and \(\theta\) goes from 0 to \(2\pi\).
These defined limits are fundamental to ensuring we're integrating over the correct area and volume.
Polar coordinates for double integrals
Using polar coordinates can significantly simplify the evaluation of double integrals, especially when the region of integration is circular or radial in nature.
  • In our exercise, converting to polar coordinates changes each point in the xy-plane from Cartesian \((x, y)\) to polar \((r, \theta)\).
  • This helps simplify the double integral because we integrate with respect to \(r\) and \(\theta\), which tidily encapsulate the circular nature of our region \(D\).
  • The differential area element \(dA\) becomes \(r \, dr \, d\theta\).
The conversion not only aligns more closely with the geometry of our region but often results in integrals that are easier to solve, both conceptually and computationally.

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Most popular questions from this chapter

Lame ovals have been consistently used by designers and architects. For instance, Gerald Robinson, a Canadian architect, has designed a parking garage in a shopping center in Peterborough, Ontario, in the shape of a superellipse of the equation \(\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=1\) with \(\frac{a}{b}=\frac{9}{7}\) and \(n=e .\) Use a CAS to find an approximation of the area of the parking garage in the case \(a=900\) yards, \(b=700\) yards, and \(n=2.72\) yards.

Find the volume of the solid \(E\) that lies under the plane \(2 x+y+z=8\) and whose projection onto the \(x y-\) plane is bounded by \(x=\sin ^{-1} y, y=0,\) and \(x=\frac{\pi}{2}\)

Let \(Q\) be the solid unit hemisphere. Find the mass of the solid if its density \(\rho\) is proportional to the distance of an arbitrary point of \(Q\) to the origin.

\([\mathbf{T}]\) a. Evaluate the integral \(\iiint_{E} e^{\sqrt{x^{2}+y^{2}+z^{2}}} d V\) where \(E\) is bounded by the spheres \(4 x^{2}+4 y^{2}+4 z^{2}=1\) and \(x^{2}+y^{2}+z^{2}=1\) b. Use a CAS to find an approximation of the previous integral. Round your answer to two decimal places.

In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(y\) -axis, and origin, respectively. \(R\) is the triangular region with vertices \((0,0),(0,3),\) and \((6,0) ; \rho(x, y)=x y\)
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