Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(7-14\), use cylindrical coordinates to find the indicated quantity. Center of mass of the homogeneous solid inside \(x^{2}+y^{2}=4\), outside \(x^{2}+y^{2}=1\), below \(z=12-x^{2}-y^{2}\), and above \(z=0\)

Step-by-Step Solution

Answer

The center of mass is at (0, 0, 0.890).

Step 1: Define the Region in Cylindrical Coordinates

Start by converting the given conditions to cylindrical coordinates. The equations \(x^2+y^2=4\) and \(x^2+y^2=1\) suggest using cylindrical coordinates where \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\). In cylindrical coordinates, these become \(r=2\) and \(r=1\), respectively. Thus, the region is bounded by \(1 \leq r \leq 2\), \(0 \leq \theta < 2\pi\), and \(0 \leq z \leq 12-r^2\).

Step 2: Setup the Integral for Mass

The mass of the solid can be found using the triple integral in cylindrical coordinates: \(m = \int_{0}^{2\pi} \int_{1}^{2} \int_{0}^{12-r^2} r \, dz \, dr \, d\theta\). We use \(r\) in the integrand because it accounts for the Jacobian of the transformation from Cartesian to cylindrical coordinates.

Step 3: Integrate with Respect to \(z\)

First, integrate with respect to \(z\):\[ \int_{0}^{12-r^2} r \, dz = r[z]_{0}^{12-r^2} = r(12-r^2) \].

Step 4: Integrate with Respect to \(r\)

Next, integrate with respect to \(r\):\[ \int_{1}^{2} r(12-r^2) \, dr = \int_{1}^{2} (12r-r^3) \, dr \].Calculate:\[ \left[ 6r^2 - \frac{r^4}{4} \right]_{1}^{2} = (24-4) - (6-\frac{1}{4}) = 20 - (6 - 0.25) = 14.25 \].

Step 5: Integrate with Respect to \(\theta\)

Finally, integrate with respect to \(\theta\):\[ \int_{0}^{2\pi} 14.25 \, d\theta = 14.25[\theta]_{0}^{2\pi} = 14.25(2\pi) = 28.5\pi \].This value, \(28.5\pi\), is the mass of the solid.

Step 6: Calculate the Center of Mass Coordinates

The center of mass coordinates \((\bar{x}, \bar{y}, \bar{z})\) are calculated as:- \(\bar{x} = \frac{1}{m} \int_{0}^{2\pi} \int_{1}^{2} \int_{0}^{12-r^2} r^2 \cos\theta \, dz \, dr \, d\theta = 0\) due to symmetry.- \(\bar{y} = \frac{1}{m} \int_{0}^{2\pi} \int_{1}^{2} \int_{0}^{12-r^2} r^2 \sin\theta \, dz \, dr \, d\theta = 0\) due to symmetry.- \(\bar{z} = \frac{1}{m} \int_{0}^{2\pi} \int_{1}^{2} \int_{0}^{12-r^2} zr \, dz \, dr \, d\theta \).

Step 7: Calculate \(\bar{z}\) Simplified Expression

First, integrate the \(z\) part:\[ \int_{0}^{12-r^2} zr \, dz = \frac{r(12-r^2)^2}{2} \]. Now integrate with respect to \(r\):\[ \int_{1}^{2} \frac{r(12-r^2)^2}{2} \, dr \].Evaluate this integral to get:\[ = \frac{1}{2} \int_{1}^{2} (144r - 48r^3 + 4r^5) \, dr \].

Step 8: Evaluate \(\bar{z}\) Integral

Perform the definite integrals:\[ \frac{1}{2} \left[ 72r^2 - 12r^4 + \frac{2r^6}{3} \right]_{1}^{2} = \frac{1}{2}(288 - 192 + \frac{128}{3} - 72 + 12 - \frac{2}{3}) = \frac{240}{3} = 80 \].Thus, \(\bar{z} = \frac{80}{28.5\pi} = \frac{160}{57\pi} \approx 0.890\).

Step 9: Conclusion of Center of Mass

Finally, we find that the center of mass is located at \((0, 0, \bar{z})\) where \(\bar{z} \approx 0.890\). This gives us the coordinates for the center of mass under the region described.

Key Concepts

Center of Mass

The center of mass is a crucial concept in physics and engineering, as it's the point at which the mass of a system is concentrated. In the context of this exercise, we're dealing with a solid region that needs its center of mass located. This point is represented by coordinates

• \(\bar{x}\)
• \(\bar{y}\)
• \(\bar{z}\)

In problems involving symmetrical objects like this solid, symmetry greatly simplifies the search for the center of mass. Due to the circular symmetry about the z-axis in our solid, we immediately deduce that both \(\bar{x}\) and \(\bar{y}\) are zero. Essentially, the center of mass lies along the z-axis, thus reducing our task to just finding \(\bar{z}\). This is achieved by evaluating a triple integral representing the average position in the z-dimension, weighted by density.

Triple Integral

To solve for the center of mass, we utilize the concept of a triple integral. This tool is invaluable in computing quantities over three-dimensional regions, such as

• the total mass
• centroids
• and centers of mass

In this exercise, the triple integral is set up in cylindrical coordinates to conveniently accommodate the geometry of the problem. It involves integrating over

• the angular coordinate (\(\theta\))
• the radial coordinate (\(r\))
• and the height (\(z\))

The overall structure of the integral is\[m = \int_{0}^{2\pi} \int_{1}^{2} \int_{0}^{12-r^2} r \ dz \ dr \ d\theta\]The integration is broken down into steps, with operation typically done inside-out:

• integrate with respect to \(z\) first
• followed by \(r\)
• and finally \(\theta\)

This sequential integration mirrors the structure of the solid, starting from its bottom and working upwards.

Symmetry in Cylindrical Coordinates

Symmetry plays a pivotal role in simplifying calculations in cylindrical coordinates, especially when considering shapes and regions with rotational symmetry. Cylindrical coordinates, defined by

• radius \(r\)
• angle \(\theta\)
• and height \(z\)

are perfectly suited for regions that exhibit symmetry around an axis, like circular solids. In this problem, the solid is symmetric about the z-axis, which allows for the simplification of finding the center of mass.
The circular symmetry means that contributions to the center of mass from \(\cos\theta\) and \(\sin\theta\) terms cancel out over the entire range of \(\theta\). Thus, both \(\bar{x}\) and \(\bar{y}\) are zero due to the symmetry. This drastically reduces the complexity of the mathematics involved, as it allows focusing the computational effort solely on the \(\bar{z}\) component.

Jacobian in Cylindrical Coordinates

In transforming from Cartesian coordinates to cylindrical coordinates for triple integration, the Jacobian expresses how volume scales under this transformation. Specifically, the Jacobian for cylindrical coordinates is \(r\), and it is crucial to include this factor when setting up integrals over a radial geometry.
The inclusion of the Jacobian ensures the integral accounts for the non-uniform scaling of space in cylindrical coordinates. The radial coordinate \(r\) changes more rapidly than Cartesian counterparts, particularly near the origin. This adjustment is necessary because each infinitesimal element

• increases in size with increasing radial distance
• ensuring proper weight is given to regions further from the axis

In our exercise, this results in the integral structure \(\int r \, dz \, dr \, d\theta\), and neglecting the Jacobian would misrepresent the physical reality of the space the solid occupies. The Jacobian thus plays a vital role in accurately representing the problem in cylindrical coordinates.