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Chapter 15: Problem 70

Centroid Find the centroid of the solid bounded above by the sphere \(\rho=a\) and below by the cone \(\phi=\pi / 4 .\)

Short Answer

Expert verified
The centroid is at \( (0, 0, \frac{a(3 - 2\sqrt{2})}{3(2 - \sqrt{2})}) \)."

Step by step solution

01

Understand the Problem

We are given a sphere with radius \( \rho = a \) and a cone with angle \( \phi = \frac{\pi}{4} \). The task is to find the centroid (center of mass) of the solid region that lies between these two surfaces.
02

Identify Region of Integration

The solid is symmetric about the z-axis. The radial limits are from the cone \( \phi = \frac{\pi}{4} \) to the sphere \( \rho = a \). In spherical coordinates this translates to angle range \( 0 \leq \theta < 2\pi \), \( \frac{\pi}{4} \leq \phi \leq \frac{\pi}{2} \), and \( 0 \leq \rho \leq a \).
03

Set Up Centroid Formulas

The centroid \( ( \bar{x}, \bar{y}, \bar{z} ) \) can be calculated using the formulas: \( \bar{x} = \frac{1}{V} \int x dV \), \( \bar{y} = \frac{1}{V} \int y dV \), \( \bar{z} = \frac{1}{V} \int z dV \), where \( V \) is the volume of the region.
04

Calculate the Volume

Compute the volume of the solid using the integral:\[ V = \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^a \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \].Evaluating this gives:\[ V = \frac{\pi a^3}{3}(2 - \sqrt{2}) \].
05

Calculate \( \bar{z} \)

Since the solid is symmetric about the z-axis, \( \bar{x} = 0 \) and \( \bar{y} = 0 \). Calculate \( \bar{z} \) using the integral:\[ \bar{z} = \frac{1}{V} \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^a (\rho \cos\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \].Evaluating this integral, we find:\[ \bar{z} = \frac{a(3 - 2\sqrt{2})}{3(2 - \sqrt{2})} \].
06

Conclusion

Therefore, the centroid of the solid is located at \( (\bar{x}, \bar{y}, \bar{z}) = (0, 0, \frac{a(3 - 2\sqrt{2})}{3(2 - \sqrt{2})}) \).

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Key Concepts

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

Spherical Coordinates in Calculus
Spherical coordinates are a three-dimensional coordinate system that extend the two-dimensional polar coordinates into three dimensions. They work great for problems involving spheres or hemispheres.
In spherical coordinates, each point in space is determined by three values:
  • \( \rho \) (rho): the distance from the origin to the point.
  • \( \phi \) (phi): the angle between the positive z-axis and the line segment from the origin to the point, known as the polar angle.
  • \( \theta \) (theta): the angle from the positive x-axis to the projection of the line segment onto the xy-plane, known as the azimuthal angle.
For this exercise, understanding these coordinates is crucial because the region of integration is defined in terms of \( \phi \) and \( \rho \). The sphere's equation is \( \rho = a \), constant for the sphere's surface, and the cone is defined by \( \phi = \frac{\pi}{4} \), the angle from the z-axis at which the cone is oriented.
Volume Integration using Calculus
In the context of this exercise, volume integration is used to find the volume of the solid region lying between the cone and the sphere.
The integral used represents the sum of tiny volume elements:\[ V = \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^a \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \].
Here's how this kind of integration works step-by-step:
  • \( \rho^2 \sin \phi \) is the Jacobian determinant for spherical coordinates, ensuring each volume element is properly scaled.
  • Limits for \( \rho \) ensure the integration moves radially outward from the origin.
  • \( \phi \) and \( \theta \) cover the angles as defined, helping encompass the full region between the cone and the sphere.
Evaluating this triple integral aids in calculating the volume of more complex shapes in spherical coordinates by integrating the density over the entire solid.
Center of Mass and Centroids
The center of mass is the average position of a mass distribution in space, while the centroid is a similar concept but for a geometric shape, not necessarily linked with mass. For symmetric, uniform density objects, its center often coincides with the centroid.
In this example, the solid is symmetric along the z-axis, allowing us to assume \( \bar{x} = 0 \) and \( \bar{y} = 0 \). The challenge is to find \( \bar{z} \), the height of the center of the solid from the base.
The integral used for \( \bar{z} \) is:
  • \( \bar{z} = \frac{1}{V} \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^a (\rho \cos\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \).
This considers how each volume slice contributes vertically. Solving it gives the centroid's vertical position, taking advantage of symmetry to simplify calculations.

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Most popular questions from this chapter

The Parallel Axis Theorem Let \(L_{\mathrm{cm}}\) be a line through the center of mass of a body of mass \(m\) and let \(L\) be a parallel line \(h\) units away from \(L_{\mathrm{cm}} .\) The Parallel Axis Theorem says the moments of inertia \(I_{\mathrm{cm}}\) and \(I_{L}\) of the body about \(L_{\mathrm{cm}}\) and \(L\) satisfy the equation $$I_{L}=I_{\mathrm{c.m.}}+m h^{2}$$ As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known.\ Proof of the Parallel Axis Theorem a. Show that the first moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the \(y z\) -plane. What does the formula \(\overline{x}=M_{y z} / M\) then tell you? b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{c . m .}\) along the \(z\) -axis and the line \(L\) perpendicular to the \(x y\) -plane at the point \((h, 0,0)\) . Let \(D\) be the region of space occupied by the body. Then, in the notation of the figure, $$I_{L}=\iiint_{D}|\mathbf{v}-h \mathbf{i}|^{2} d m$$ Expand the integrand in this integral and complete the proof.

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals in Exercises \(27-30\) . $$ \int_{0}^{2} \int_{-\pi}^{0} \int_{\pi / 4}^{\pi / 2} \rho^{3} \sin 2 \phi d \phi d \theta d \rho $$

Find the average value of the function \(f(r, \theta, z)=r\) over the region bounded by the cylinder \(r=1\) between the planes \(z=-1\) and \(z=1\) .

Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry three-eighths of the way from the base toward the top, show, by transforming the appropriate integrals, that the center of mass of a solid semiellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\left(z^{2} / c^{2}\right) \leq 1, \quad z \geq 0\) lies on the \(z\) -axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)

In Exercises \(15-20,\) set up the iterated integral for evaluating \(\int \int \int_{D} f(r, \theta, z) d z r d r d \theta\) over the given region \(D .\) $$ \begin{array}{l}{D \text { is the right circular cylinder whose base is the circle } r=2 \sin \theta} \\ {\text { in the } x y \text { -plane and whose top lies in the plane } z=4-y}\end{array} $$
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