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Magazine Chapter 14: Problem 9
Find the centroid of the region. Use symmetry wherever possible to reduce calculations. The solid region bounded above by the plane \(z=1\) and below by the upper nappe of the cone \(z^{2}=9 x^{2}+9 y^{2}\)
Short Answer
Expert verified
The centroid is \((0, 0, \frac{3}{4})\).
Step by step solution
01
Analyze the Problem
The solid region is bounded by the plane \(z=1\) and the upper nappe of the cone defined by \(z^2 = 9x^2 + 9y^2\). This can be simplified to \(z = 3\sqrt{x^2+y^2}\). We are tasked with finding the centroid of this region using symmetry where possible.
02
Identify Symmetry
The region is symmetric around the z-axis and the x-y plane, suggesting that the x and y coordinates of the centroid will be zero due to symmetry. Therefore, the challenge reduces to finding the z-coordinate of the centroid.
03
Set up the Volume Integration
The limits of integration for the region will be set in cylindrical coordinates: \(z\) varies from \(3r\) (the cone) to \(1\) (the plane), and \(r\) varies from \(0\) to \(1/3\) because where a circle intersects the plane \(z=1\) and the cone's equation, \(1=3r\) yields \(r=1/3\). Integrate \(\int_0^{2\pi}\int_0^{1/3}\int_{3r}^{1} r\,\text{d}z\,\text{d}r\,\text{d}\theta\).
04
Solve Volume Integral
Evaluate the integral step-by-step. First, integrate over \(z\): \[\int_{3r}^{1} r\,\text{d}z = r[z]_{3r}^{1} = r(1 - 3r)\].Next, integrate over \(r\):\[\int_0^{1/3} r(1 - 3r) \,\text{d}r = \left[\frac{r^2}{2} - r^3 \right]_0^{1/3} = \frac{1}{18} - \frac{1}{27} = \frac{1}{54}\].Finally, integrate over \(\theta\): \[\int_0^{2\pi} \frac{1}{54} \,\text{d}\theta = \frac{2\pi}{54} = \frac{\pi}{27}\].The total volume is \(\frac{\pi}{27}\).
05
Find Centroid's Z-coordinate
The z-coordinate of the centroid (\(\bar{z}\)) is found using the formula: \[ \bar{z} = \frac{1}{V}\int \int \int z \,dV \].Calculate this by setting up the integral \(\int_0^{2\pi}\int_0^{1/3}\int_{3r}^{1} zr\,\text{d}z\,\text{d}r\,\text{d}\theta\).
06
Solve for \(\bar{z}\)
Evaluate the integral for the z-coordinate, starting with \(z\):\[\int_{3r}^{1} z rz \,\text{d}z = \frac{r}{2}[z^2]_{3r}^{1} = \frac{r}{2}((1)^2 - (3r)^2) = \frac{r}{2}(1 - 9r^2)\].Now, integrate over \(r\):\[\int_0^{1/3} \frac{r(1 - 9r^2)}{2} \,\text{d}r = \left[\frac{r^2}{4} - \frac{9r^4}{8} \right]_0^{1/3} = \frac{1}{36} - \frac{1}{72} = \frac{1}{72}\].Next, integrate over \(\theta\): \[\int_0^{2\pi} \frac{1}{72} \,\text{d}\theta = \frac{1}{72} \times 2\pi = \frac{\pi}{36}\].Thus, \(\bar{z} = \frac{\pi/36}{\pi/27} = \frac{27}{36} = \frac{3}{4}\).
07
Assemble the Centroid
With symmetry, the centroid of the region is \((\bar{x}, \bar{y}, \bar{z}) = (0, 0, \frac{3}{4})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Centroid of a solid
Finding the centroid of a three-dimensional solid is similar to finding the center of mass, assuming a uniform density. The centroid is essentially the average position of all the points in a solid.
- For symmetrical solids, the centroid often lies along axes of symmetry.
- This reduces calculations, as some coordinates of the centroid can be determined by symmetry alone.
Symmetry in integration
Symmetry is a powerful tool in integration that can simplify complex calculations. When a region in space is symmetric about certain axes or planes, this symmetry can be exploited.
- In our region, symmetry about the z-axis means that the x and y coordinates of the centroid are zero.
- This spares us from separate integrations for x and y directions.
Cylindrical coordinates
Cylindrical coordinates are particularly useful when the geometry of a problem involves cylinders or circular symmetry.
- The coordinates are represented as \(r, \, \theta, \, z\), where r is the radial distance, \(\theta\) is the angular coordinate, and z is the height.
- They simplify the integration for solids that have circular components.
Volume integration
Volume integration helps to determine volumes, centroids, and moments of inertia of solid bodies by integrating over a volume.
- For our exercise, we're using triple integrals which give us the total volume and the centroid's z-coordinate.
- By integrating the function \(r - 3r^2\) over the region, we find the volume, \(\frac{\pi}{27}\).
