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The solid \(Q=\left\\{(x, y, z) \mid 0 \leq x^{2}+y^{2} \leq 16, x \geq 0, y \geq 0,0 \leq z \leq x\right\\}\) has the density \(\rho(x, y, z)=k\). Show that the moment \(M_{x y}\) about the \(x y\) -plane is half of the moment \(M_{y z}\) about the \(y z\) -plane.

Short Answer

Expert verified
The moment about the \(xy\)-plane \(M_{xy}\) is half of the moment about the \(yz\)-plane \(M_{yz}\).

Step by step solution

01

Define the Moments

The moment about the \(xy\)-plane, \(M_{xy}\), for a density \(\rho(x, y, z)\) is given by \[M_{xy} = \iiint_{Q} z \cdot \rho(x, y, z) \, dV.\] The moment about the \(yz\)-plane, \(M_{yz}\), is given by \[M_{yz} = \iiint_{Q} x \cdot \rho(x, y, z) \, dV.\] Here, the domain \(Q\) is defined by \(0 \leq x^{2} + y^{2} \leq 16\), \(x \geq 0, y \geq 0, 0 \leq z \leq x\), and \(\rho(x, y, z) = k\).
02

Set Up the Limits of Integration

Convert the boundary conditions of \(Q\) to cylindrical coordinates: \(x = r\cos\theta\), \(y = r\sin\theta\), \(z = z\), where \(0 \leq r \leq 4\), \(0 \leq \theta \leq \frac{\pi}{2}\), and \(0 \leq z \leq r\cos\theta\). The differential volume element is \(dV = r\,dz\,dr\,d\theta\).
03

Calculate \(M_{xy}\)

Substitute \(\rho(x, y, z) = k\) and integrate over \(z\), \(r\), and \(\theta\):\[M_{xy} = k \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} \int_{0}^{r\cos\theta} z \cdot r \, dz \, dr \, d\theta.\]Evaluate the inner integral with respect to \(z\):\[\int_{0}^{r\cos\theta} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{r\cos\theta} = \frac{(r\cos\theta)^2}{2}.\]This becomes:\[M_{xy} = k \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} \frac{r^3\cos^2\theta}{2} \, dr \, d\theta = \frac{k}{2} \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r^3\cos^2\theta \, dr \, d\theta.\]
04

Calculate \(M_{yz}\)

Substitute \(\rho(x, y, z) = k\) and set up the triple integral: \[M_{yz} = k \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} \int_{0}^{r\cos\theta} (r\cos\theta) \cdot r \, dz \, dr \, d\theta.\]The \(z\) integral simplifies because it's constant over \(z\):\[\int_{0}^{r\cos\theta} r^2\cos\theta \, dz = r^2\cos\theta \cdot r\cos\theta = r^3\cos^2\theta.\]Thus:\[M_{yz} = k \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r^3\cos^2\theta \, dr \, d\theta.\]
05

Evaluate the Integrals

Notice that \(M_{yz}\) and \(M_{xy}\) integrals are similar except for a factor:\[M_{xy} = \frac{k}{2} \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r^3\cos^2\theta \, dr \, d\theta,\]\[M_{yz} = k \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r^3\cos^2\theta \, dr \, d\theta.\]This directly shows:\[M_{xy} = \frac{1}{2} M_{yz}.\] Thus, the moment about the \(xy\)-plane is half of the moment about the \(yz\)-plane.

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Key Concepts

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

Cylindrical Coordinates
In mathematics, we often use cylindrical coordinates to simplify the integration process for objects that have cylindrical symmetry, such as cylinders and cones. Cylindrical coordinates are expressed in terms of three parameters:
  • **Radius (\(r\))**: This is the distance from the point to the origin in the xy-plane. It's similar to the radius in polar coordinates.
  • **Angle (\(\theta\))**: This is the angle between the positive x-axis and the line connecting the origin to the point in the xy-plane. It's also akin to the angle used in polar coordinates.
  • **Height (\(z\))**: This is just the standard z-coordinate used in Cartesian coordinates.
In our specific problem, cylindrical coordinates greatly simplify the integration over the solid Q. The transformation from Cartesian coordinates to cylindrical coordinates is done using the formulas:
  • \(x = r\cos\theta\)
  • \(y = r\sin\theta\)
With these transformations, we can easily define boundaries for integrals that describe objects extending out from the origin across angles and heights.
Triple Integrals
Triple integrals are a powerful tool for calculating various physical properties over three-dimensional regions. They extend double integrals from two-dimensional areas to three-dimensional volumes, allowing us to find quantities like mass, center of mass, and moments of inertia.In practice, a triple integral is often of the form:\[ \iiint_Q f(x, y, z) \, dV \]Here, \(f(x, y, z)\) is a function that may represent density, temperature, or some other property that varies at each point within the volume \(Q\). The differential volume element \(dV\) represents a tiny part of the total volume over which we are integrating the function.In the context of our original exercise, we use a triple integral to calculate the moments \(M_{xy}\) and \(M_{yz}\). These moments are essentially weighted averages over the volume \(Q\), where the 'weights' are the coordinates \(z\) and \(x\) respectively. Integrating over each coordinate direction allows us to analyze how mass and distance interact throughout the solid.
Volume Integration
Volume integration is integral calculus's method extended to three dimensions, helping us quantify the total effect of a function within a space. When calculating moments or other properties over a solid, we use volume integration to account for both shape and size.In volume integration, the triple integral mentioned previously is essential. The integral can be broken down into three simpler integrals performed over the limits of each variable. In the cylindrical coordinates used for our exercise:
  • First, integrate with respect to \(z\), which is often the 'height'.
  • Next, move to the radial component \(r\), which spans from the center outward.
  • Finally, cover the angular component \(\theta\), which accounts for rotation around the z-axis.
This sequence ensures that each integral captures part of the three-dimensional nature of the solid. This approach helps us understand complex systems, like the solid in the problem, by considering how each section contributes to the whole. The key to mastering volume integration is getting comfortable with setting these limits and understanding the geometry of the problem.

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Most popular questions from this chapter

Determine the average value of the function \(f(x, y)=x^{2}+y^{2}\) over the region \(D\) bounded by the polar curve \(r=\cos 2 \theta\), where \(-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}\) (see the following graph)

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