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Chapter 14: Problem 8

Find the centroid of the region. Use symmetry wherever possible to reduce calculations. The solid region bounded below by the paraboloid \(z=\) \(4 x^{2}+4 y^{2}\) and above by the plane \(z=2\)

Short Answer

Expert verified
The centroid is \((0, 0, \frac{4}{3})\).

Step by step solution

01

Understand the Region

We are considering the volume bounded below by the paraboloid \(z = 4x^2 + 4y^2\) and above by the plane \(z = 2\). The intersection of these surfaces will form the boundary in the xy-plane.
02

Find the Intersection Curve

To find the intersection curve in the xy-plane, we equate the expressions for \(z\): \(4x^2 + 4y^2 = 2\). Simplifying, we get \(x^2 + y^2 = \frac{1}{2}\) which represents a circle of radius \(\frac{1}{\sqrt{2}}\).
03

Set Up the Integral for Volume

The volume \(V\) of the region is found using a double integral. Since the region is circular, using polar coordinates is convenient. The limits for \(r\) go from 0 to \(\frac{1}{\sqrt{2}}\), and the limits for \(\theta\) go from 0 to \(2\pi\). The volume integral is \( V = \int_0^{2\pi} \int_0^{\frac{1}{\sqrt{2}}} (2 - 4r^2)r \, dr \, d\theta \).
04

Calculate the Volume

Evaluate the inner integral: \(\int_0^{\frac{1}{\sqrt{2}}} (2r - 4r^3) \, dr\). This yields after integration \(\left[ r^2 - r^4 \right]_0^{\frac{1}{\sqrt{2}}} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \). Then integrate with respect to \(\theta\): \(\int_0^{2\pi} \frac{1}{4} \, d\theta = \frac{1}{4}(2\pi) = \frac{\pi}{2} \). Thus, \(V = \frac{\pi}{2}\).
05

Find the Centroid Coordinates

By symmetry, the centroid lies on the z-axis due to the region being symmetrical about it. Thus \(\bar{x} = \bar{y} = 0\). For \(\bar{z}\), use \( \bar{z} = \frac{1}{V} \int \int \int z \, dV \). For polar coordinates, this becomes \(\bar{z} = \frac{1}{V} \int_0^{2\pi} \int_0^{\frac{1}{\sqrt{2}}} \int_{4r^2}^{2} z \, r \, dz \, dr \, d\theta \).
06

Perform the Triple Integral for \(\bar{z}\)

Evaluate the inner integral \( \int_{4r^2}^{2} z \, dz = \left[ \frac{z^2}{2} \right]_{4r^2}^{2} = \frac{4}{2} - \frac{(4r^2)^2}{2} = 2 - 8r^4 \). Set up the remaining integrals: \( \int_0^{2\pi} \int_0^{\frac{1}{\sqrt{2}}} (2r - 8r^5) \, dr \, d\theta \).
07

Calculate and Finalize \(\bar{z}\)

Calculate the integral \( \int_0^{\frac{1}{\sqrt{2}}} (2r - 8r^5) \, dr = \left[ r^2 - \frac{8r^6}{6} \right]_0^{\frac{1}{\sqrt{2}}} = \frac{1}{2} - \frac{8}{6} \cdot \frac{1}{8} = \frac{1}{3} \). Integrate with respect to \(\theta\), giving \( \frac{1}{3}(2\pi) = \frac{2\pi}{3} \). Then \(\bar{z} = \frac{2\pi/3}{\pi/2} = \frac{4}{3}\). Thus, the centroid is \((0, 0, \frac{4}{3})\).

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Key Concepts

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

Symmetry
When calculating the centroid of a symmetrical object, symmetry can greatly simplify the process. Symmetry means that the object is identical on both sides of a dividing line or plane. In this exercise, the region is symmetrical around the z-axis. This simplifies calculations because:
  • We know that the x and y coordinates of the centroid will be zero since the region is perfectly balanced around the z-axis.
  • This symmetry reduces the complexity of the calculation, as it eliminates the need to perform additional integrals for \( \bar{x} \) and \( \bar{y} \).
Understanding symmetry allows you to reduce the work needed to find the centroid, making calculations more efficient and less error-prone.
Paraboloid
A paraboloid is a three-dimensional surface described by a quadratic equation. In this problem, the paraboloid is given by \(z = 4x^2 + 4y^2\). This surface is bowl-shaped, opening upward, and is symmetric about the z-axis. Key characteristics include:
  • The vertex of this paraboloid is located at the origin (0,0,0).
  • It extends infinitely in the z-direction but is bounded by the plane \(z=2\).
The paraboloid's equation defines the lower bound of our region, while the plane \(z=2\) forms the upper boundary. This setup creates a finite, circular region, due to the circular boundary in the xy-plane. This understanding helps set the limits for integration.
Polar Coordinates
Polar coordinates are used to simplify calculations involving circular regions or when dealing with problems of symmetry. Instead of thinking in terms of x and y, polar coordinates use:
  • Radius (\(r\)), which is the distance from the origin.
  • Angle (\(\theta\)), which is the angle from the positive x-axis.
This approach is particularly useful when the region is circular, as it becomes straightforward to describe boundaries. In this exercise:
  • The conversion of the circle \(x^2 + y^2 = \frac{1}{2}\) is expressed as \(r = \frac{1}{\sqrt{2}}\).
  • The limits for \(r\) and \(\theta\) are 0 to \(\frac{1}{\sqrt{2}}\) and 0 to \(2\pi\), respectively.
This substitution simplifies the integration, because the region can be integrated more naturally in terms of its size and shape.
Triple Integral
A triple integral enables us to compute the volume of a three-dimensional region or to find average values, like the centroid. In three-dimensional space, it's expressed as \(\int \int \int dV\). The steps for using triple integrals include:
  • Setting limits of integration for each coordinate based on the region's boundaries.
  • Integrating step-by-step, starting with innermost to the outermost function.
Given our region between the paraboloid and plane:
  • Integrate first with respect to \(z\), then \(r\), and finally \(\theta\).
  • Use the volume and \(\bar{z}\) formulas to find relevant measures, evaluating each integral accordingly.
The triple integral performs an essential function by summing infinitesimally small volumes, providing a complete measure of the region's properties.

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Most popular questions from this chapter

Approximate the integral with the help of a computer algebra system. \(\int_{0}^{1} \int_{0}^{x} \sqrt{x^{4}+y^{4}} d y d x \quad\) (Hint: How does the integral $$ \begin{aligned} &\int_{0}^{1} \int_{0}^{x} \sqrt{x^{4}+y^{4}} d y d x \text { compare with } \\\ &\left.\int_{0}^{1} \int_{0}^{1} \sqrt{x^{4}+y^{4}} d y d x ?\right) \end{aligned} $$ Let \(R\) be a vertically or horizontally simple region in the \(x y\) plane, and let \(f\) be continuous on \(R .\) The average value (or mean value) \(f_{\text {av }}\) of \(f\) on \(R\) is given by $$ f_{\mathrm{av}}=\frac{1}{\text { area of } R} \iint_{R} f(x, y) d A $$

Find the area \(A\) of the region in the \(x y\) plane by the methods of this section. The region in the first quadrant that is bounded by the graph of \(x=2-y^{2}\) and by the lines \(x=0\) and \(x=y\)

Find the volume \(V\) of the region, using the methods of this section. The solid region bounded above by the parabolic sheet \(z=1-x^{2}\), below by the \(x y\) plane, and on the sides by the planes \(y=-1\) and \(y=2\).

Evaluate the integral by using the given transformation. \(\iint_{R} \frac{y}{x-3 y} d A\), where \(R\) is the region bounded by the lines \(y=1, y=\frac{1}{4} x\), and \(x-3 y=e ;\) let \(x=3 u+v, y=u\)

Reverse the order of integration and evaluate the resulting integral. \(\int_{0}^{\sqrt{\pi / 2}} \int_{y}^{\sqrt{\pi / 2}} y^{2} \sin x^{2} d x d y\) (Hint: After changing the order of integration, make the substitution \(u=x^{2}\).)
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