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Chapter 1: Problem 7

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 consists of two circles with radius 2 at \(z = \sqrt{5}\) and \(z = -\sqrt{5}\).

Step by step solution

01

Understand the equations

The first equation, \(x^{2} + y^{2} + z^{2} = 9\), represents a sphere with radius 3 centered at the origin. The second equation, \(x^{2} + y^{2} = 4\), represents a cylinder with radius 2 centered along the z-axis.
02

Combine the equations

To find the intersection, set the equation of the cylinder \(x^{2} + y^{2} = 4\) into the equation of the sphere \(x^{2} + y^{2} + z^{2} = 9\). Substituting \(x^{2} + y^{2} = 4\) into \(x^{2} + y^{2} + z^{2} = 9\), we get \(4 + z^{2} = 9\).
03

Simplify the equation

Solve the equation \(4 + z^{2} = 9\) to find the values of \(z\). Subtract 4 from both sides to obtain \(z^{2} = 5\).
04

Solve for z

Take the square root of both sides to find \(z\). We get \(z = \pm \sqrt{5}\).
05

Describe the intersection curve

The intersection of the sphere and the cylinder occurs at the two circles. These circles are parallel to the xy-plane at \(z = \sqrt{5}\) and \(z = -\sqrt{5}\). The radius of each circle is 2 (from the cylinder equation \(x^{2} + y^{2} = 4\)).

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Key Concepts

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

intersection of surfaces
The concept of intersection of surfaces in 3D geometry involves finding the common set of points that satisfy the equations of both surfaces. In this exercise, we are dealing with two surfaces: a sphere and a cylinder. The sphere is defined by the equation \(x^{2} + y^{2} + z^{2} = 9\), meaning it includes all points in space that are at a distance of 3 from the origin. The cylinder is defined by \(x^{2} + y^{2} = 4\), indicating it consists of circles of radius 2 centered along the z-axis and extending infinitely in both directions.
sphere and cylinder intersection
To find the intersection between a sphere and a cylinder, we substitute the equation of the cylinder into the equation of the sphere. In this case, we take \(x^{2} + y^{2} = 4\) from the cylinder equation and substitute it into \(x^{2} + y^{2} + z^{2} = 9\), the equation of the sphere. This gives us: \(4 + z^{2} = 9\). To find the z-values where the two surfaces intersect, we simplify this to \(z^{2} = 5\) and solve for \(z\), yielding \(z = \pm \sqrt{5}\). The intersection points lie on two circles, one at \(z = \sqrt{5}\) and one at \(z = -\sqrt{5}\). These circles are parallel to the xy-plane and have a radius of 2, as dictated by the cylinder equation.
3D geometry
Understanding these geometric shapes and their intersections in three-dimensional space is crucial in vector calculus and 3D geometry. The sphere and the cylinder each form fundamental 3D surfaces. In vector calculus, we often use parametric equations or vector functions to express points on these surfaces. Recognizing that the intersection forms circles shows how different surfaces interact in three dimensions. By substituting one equation into another, we simplify and solve for the remaining variable, making the problem more manageable. This skill is essential for many applications in physics, engineering, and computer graphics.

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Most popular questions from this chapter

For Exercises \(1-4,\) determine if the given equation deseribes a sphere. If so, find its radius and center. $$ x^{2}+y^{2}-z^{2}+12 x+2 y-4 z+32=0 $$

Show that the hyperboloid of one sheet is a doubly ruled surface, i.e. each point on the surface is on two lines lying entirely on the surface. Hint: Write equation (1.35) as \(\frac{x^{2}}{a^{2}}-\frac{z^{2}}{e^{2}}=1-\frac{y^{2}}{b^{2}},\) factor each side. Recall that two planes intersect in a line)

Find the trace of the hyperbolic paraboloid \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=\frac{z}{c}\) in the \(x y\) -plane.

Show that \(\frac{d}{d t}(\mathbf{f} \cdot(\mathbf{g} \times \mathbf{h}))=\frac{d \mathbf{f}}{d t} \cdot(\mathbf{g} \times \mathbf{h})+\mathbf{f} \cdot\left(\frac{d \mathbf{g}}{d t} \times \mathbf{h}\right)+\mathbf{f} \cdot\left(\mathbf{g} \times \frac{d \mathbf{h}}{d t}\right) .\)

Let \(\mathbf{r}(t)\) be the position vector for a particle moving in \(\mathbb{R}^{3}\). Show that $$ \frac{d}{d t}(\mathbf{r} \mathbf{x}(\mathbf{v} \times \mathbf{r}))=\|\mathbf{r}\|^{2} \mathbf{a}+(\mathbf{r} \cdot \mathbf{v}) \mathbf{v}-\left(\|\mathbf{v}\|^{2}+\mathbf{r} \cdot \mathbf{a}\right) \mathbf{r} . $$
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