/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Find the volume of the smaller r... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 51

Find the volume of the smaller region cut from the solid sphere \(\rho \leq 2\) by the plane \(z=1\).

Short Answer

Expert verified
The volume of the smaller region is \(\frac{5}{3}\pi\).

Step by step solution

01

Understand the Problem

We need to find the volume of the smaller region of the sphere defined by \(\rho \leq 2\) when it is cut by the plane \(z = 1\). The sphere is centered at the origin and has a radius of 2, and the plane cuts the sphere horizontally at \(z = 1\).
02

Determine the Cap Height

The smaller region is a spherical cap. The height of this cap, \(h\), is the distance from the center of the sphere to the plane \(z = 1\). Since the sphere is centered at the origin, its diameter is 4, and its top is at \(z = 2\), the height is given by \(h = 2 - 1 = 1\).
03

Use the Spherical Cap Formula

The volume \(V\) of a spherical cap is given by the formula \(V = \frac{1}{3}\pi h^2 (3R - h)\), where \(h\) is the height of the cap and \(R\) is the radius of the sphere. For our cap, \(h = 1\) and \(R = 2\). Substitute these values in to find \(V\).
04

Calculate the Volume

Substituting into the formula: \[V = \frac{1}{3}\pi (1)^2 (3 \times 2 - 1) = \frac{1}{3}\pi \times 1 \times 5 = \frac{5}{3}\pi.\]The volume of the smaller region cut from the sphere is \(\frac{5}{3}\pi\).

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Key Concepts

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

Spherical Cap
A spherical cap is a section of a sphere, typically created when a sphere is intersected by a plane. Visualize a sphere, like a small rubber ball, cut into two parts by a flat plane. The upper or lower section that results from this cut is the spherical cap. In the sphere with radius 2, sliced by the plane at \(z = 1\), the top slice forms our spherical cap.

Every spherical cap has unique characteristics:
  • Base area - formed by the intersection where the plane meets the sphere.
  • Height \(h\) - this is the vertical distance from the base of the cap to the top of the sphere.
In our exercise, the height \(h\) of the spherical cap is 1. This was found by subtracting the \(z\) value of the plane from the topmost \(z\) value of the sphere, which is 2.
Volume Calculation
Calculating the volume of a spherical cap involves using a specific formula. Simply knowing that a spherical cap has certain dimensions isn't enough; we need a mathematical approach to find its volume.

The formula for finding the volume \(V\) of a spherical cap is: \[ V = \frac{1}{3}\pi h^2 (3R - h) \]Here, \(h\) represents the height of the cap, and \(R\) is the radius of the entire sphere.

For example, let's use the parameters from our original exercise:
  • Height \(h = 1\)
  • Radius \(R = 2\)
Plug these values into the formula:\[ V = \frac{1}{3}\pi (1)^2 (3 \times 2 - 1) = \frac{1}{3}\pi \times 1 \times 5 = \frac{5}{3}\pi \]Hence, the calculated volume of the spherical cap is \(\frac{5}{3}\pi\). Every piece of the formula plays a critical role in ensuring that the volume is correctly determined.
Mathematical Problem Solving
Mathematical problem solving is a skill that extends beyond simply crunching numbers; it's about understanding the components of a problem and strategically evaluating solutions. When tackling exercises like finding the volume of a spherical cap, the key steps are:

  • Understanding the Problem - Clarify what's being asked. In our case, identify the geometry: a sphere cut by a plane, forming a cap.
  • Identify Known Values - Determine the cap's height, sphere's radius, or any given variables. Here, height \(h = 1\) and radius \(R = 2\).
  • Apply the Right Formula - Use the correct formula for a spherical cap to find the volume: \(V = \frac{1}{3}\pi h^2 (3R - h)\).


Problem solving isn't just computation; it involves logical reasoning and clear analysis of the gathered information to reach an accurate solution. This approach can be applied to a wide array of mathematical challenges beyond just this sphere problem.

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Most popular questions from this chapter

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{0}^{2} \int_{0}^{4-y^{2}} y d x d y$$

The Parallel Axis Theorem Let \(L_{\mathrm{c}, \mathrm{m}}\) 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 that the moments of inertia \(I_{\mathrm{c} . \mathrm{m}}\) and \(I_{L}\) of the body about \(L_{\mathrm{c} . \mathrm{m}}\) and \(L\) satisfy the equation $$I_{L}=I_{\mathrm{cm}}+m h^{2},$$ 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. 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 \(\bar{x}=M_{y z} / M\) then tell you?)To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{\mathrm{c}, \mathrm{m} .}\) 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.

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Sketch the region of integration, reverse the order of integration, and evaluate the integral. Find the volume of the solid whose base is the region in the \(x y-\) plane that is bounded by the parabola \(y=4-x^{2}\) and the line \(y=3 x,\) while the top of the solid is bounded by the plane \(z=x+4\)

Integrate \(\quad f(x, y)=\) \(\left[\ln \left(x^{2}+y^{2}\right)\right] /\left(x^{2}+y^{2}\right)\) over the region \(1 \leq x^{2}+y^{2} \leq e^{2}\)
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