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Chapter 1: Problem 8

Find the intersection of the sphere \(x^{2}+y^{2}+z^{2}=9\) and the cylinder \(x^{2}+y^{2}=4\).

Short Answer

Expert verified
The intersection is the circles \(x^2 + y^2 = 4\) in the planes \(z = \sqrt{5}\) and \(z = -\sqrt{5}\).

Step by step solution

01

- Understand the Equations

The given sphere has the equation \(x^2 + y^2 + z^2 = 9\) and the cylinder has the equation \(x^2 + y^2 = 4\). To find their intersection, we must solve these equations simultaneously.
02

- Substitute Cylinder Equation into Sphere Equation

The cylinder's equation \(x^2 + y^2 = 4\) tells us that the intersection lies on a circle of radius 2 in the \(xy\)-plane. Substitute \(x^2 + y^2 = 4\) into the sphere equation \(x^2 + y^2 + z^2 = 9\): \[ 4 + z^2 = 9 \]
03

- Solve for z

Rearrange \[ 4 + z^2 = 9 \] to solve for \(z\): \[ z^2 = 9 - 4 = 5 \] \[ z = \pm \sqrt{5} \]
04

- Identify the Intersection

The solutions for \(z\) are \(z = \sqrt{5} \) and \(z = -\sqrt{5}\). The intersection of the sphere and cylinder will therefore be the circles \(x^2 + y^2 = 4\) lying in the planes \(z = \sqrt{5}\) and \(z = -\sqrt{5}\).

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Key Concepts

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

Sphere Equation
A sphere in three-dimensional space is a perfectly symmetrical shape where every point on its surface is equidistant from its center. The general equation for a sphere is \(x^2 + y^2 + z^2 = r^2\), where \(r\) is the radius of the sphere. In the given problem, the sphere's equation is \(x^2 + y^2 + z^2 = 9\). This tells us that the sphere is centered at the origin (0, 0, 0) and has a radius of 3 (since \( \sqrt{9} = 3\)). To understand how this sphere intersects with other shapes, like the cylinder, we need to explore their equations together.
Cylinder Equation
A cylinder is a geometric shape with straight parallel sides and a circular cross-section. The equation of a cylinder is simpler than that of a sphere. For cylinders aligned along the z-axis, the equation typically looks like \(x^2 + y^2 = r^2\), where \(r\) is the radius of the cylinder's base. In this exercise, the cylinder's equation is \(x^2 + y^2 = 4\), indicating a cylinder with a radius of 2 and an infinite height. This means the cylinder extends infinitely along the z-axis but has a circular cross-section of radius 2 in the xy-plane.
Solving Equations Simultaneously
To find the intersection of two geometric shapes, we must solve their equations simultaneously. Here, we need to find where the sphere \(x^2 + y^2 + z^2 = 9\) and the cylinder \(x^2 + y^2 = 4\) intersect.
Start by substituting the cylinder's equation \(x^2 + y^2 = 4\) into the sphere's equation \(x^2 + y^2 + z^2 = 9\). This substitution simplifies to \4 + z^2 = 9\. Then, solve for \(z\):
  • \z^2 = 9 - 4 = 5\
  • \z = \pm \sqrt{5}\
The solutions for \(z\) are \(z = \sqrt{5} \) and \(z = -\sqrt{5} \).
This means the cylinder intersects the sphere in two horizontal circles in the planes \(z = \sqrt{5}\) and \(z = -\sqrt{5}\), with each circle lying in the xy-plane.

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Most popular questions from this chapter

Show that the distance \(d\) between the points \(P_{1}\) and \(P_{2}\) with cylindrical coordinates \(\left(r_{1}, \theta_{1}, z_{1}\right)\) and \(\left(r_{2}, \theta_{z}, z_{2}\right),\) respectively, is $$ d=\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{2}-\theta_{1}\right)+\left(z_{2}-z_{1}\right)^{2}} $$

Calculate the magnitudes of the following vectors: (a) \(\mathbf{v}=(2,-1)\) (b) \(\mathbf{v}=(2,-1,0)\) (c) \(\mathbf{v}=(3,2,-2)\) (d) \(\mathbf{v}=(0,0,1)\) (e) \(\mathbf{v}=(6,4,-4)\)

Find the line of intersection (if any) of the given planes. $$ 3 x+y-5 z=0, x+2 y+z+4=0 $$

For Exercises \(1-6,\) calculate \(\mathbf{v} \times \mathbf{w}\). $$ \mathbf{v}=\mathbf{i}, \mathbf{w}=3 \mathbf{i}+2 \mathbf{j}+4 \mathbf{k} $$

Show that for \(a \neq 0,\) the equation \(\rho=2 a \sin \phi \cos \theta\) in spherical coordinates describes a sphere centered at \((a, 0,0)\) with radius \(|a|\).
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