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Chapter 11: Problem 49

Relate to cylindrical coordinates defined by \(x=r \cos \theta, y=r \sin \theta\) and \(z=z.\) Find parametric equations for the wedge in the first octant bounded by \(y=0, y=x, x^{2}+y^{2}=4, z=0\) and \(z=1.\)

Short Answer

Expert verified
The parametric equations for the wedge in cylindrical coordinates are: \(r = r\), with \(0 \leq r \leq 2\); \(胃 = 胃\), with \(0 \leq 胃 \leq 蟺/4\); and \(z = z\), with \(0 \leq z \leq 1\).

Step by step solution

01

Understand the Given Boundaries in Cartesian Coordinates

Our boundaries are defined as follows: \(y=0\), \(y=x\), \(x^{2}+y^{2}=4\), \(z=0\) and \(z=1\). Looking at the boundary equations, one can see that the wedge is in the first octant and bounded by the plane \(y=x\), the cylinder \(x^{2}+y^{2}=4\) (with radius 2) and the planes \(z=0\) and \(z=1\).
02

Conversion to Cylindrical Coordinates

Now we will use the conversion relations \(x=r \cos \theta\), \(y=r \sin \theta\) and \(z=z\) to convert our boundaries into cylindrical coordinates. The conversion leads to the following boundaries for the variables r, 胃, and z:\n\n1. For 胃: Because of the planes \(y=0\) and \(y=x\), 胃 varies from 0 to \(蟺/4\).\n2. For r: Due to the circle \(x^{2}+y^{2}=4\), r goes from 0 to 2.\n3. For z: As defined by the boundaries \(z=0\) and \(z=1\), z varies from 0 to 1.
03

Parametric Equations for the Wedge

Given the aforementioned boundaries for r, 胃, and z, we can now write the parametric equations for the wedge in cylindrical coordinates. These become \(r = r\), where \(0 \leq r \leq 2\); \(胃 = 胃\), where \(0 \leq 胃 \leq 蟺/4\); and \(z = z\), where \(0 \leq z \leq 1\).

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Key Concepts

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

Parametric Equations
Parametric equations are a way to represent curves and surfaces in mathematics and physics. They are especially helpful when dealing with complex boundaries because they allow us to describe a curve using parameters rather than being confined to a function in Cartesian coordinates. This is very useful in three-dimensional space, where objects can have more complicated shapes.
  • In the context of cylindrical coordinates, we often use parameters like radius \(r\), angle \(\theta\), and height \(z\) to describe regions or surfaces.
  • These parameters can help define boundaries and surfaces more easily than traditional Cartesian methods.
In this exercise, parametric equations help create a more versatile mathematical description of the wedge, making it straightforward to visualize and analyze.
First Octant
The first octant is a specific section of three-dimensional space where the coordinates \(x\), \(y\), and \(z\) are all positive. Think of it like being in the corner of a room where all lengths from the walls and floor outward are positive.
  • In mathematical terms, the first octant is defined as the region where \( x > 0 \), \( y > 0 \), and \( z > 0 \).
  • This area is one-eighth of the entire three-dimensional space, making it a simple yet significant domain for problems requiring bounded positive measures.
For example, in our exercise, the wedge lies entirely within this first octant. This means any parametric equations or boundaries are limited to positive values for \(x\), \(y\), and \(z\). This is critical in ensuring that all calculations and visualizations stay within the desired region.
Cartesian Coordinates
Cartesian coordinates are a way to determine a point's position in space using perpendicular axes. In three dimensions, these axes are denoted as \(x\), \(y\), and \(z\). This system is named after Ren茅 Descartes and is widely used because it simplifies complex geometric concepts.
  • Each point in space can be described using a set of three numbers \((x, y, z)\), which represent distances along the three perpendicular axes.
  • This representation makes it easy to plot points and find intersections within given boundaries or surfaces.
In our exercise, the wedge is bounded by equations in Cartesian coordinates such as \(y = 0\), \(y = x\), and \(x^2 + y^2 = 4\). These equations define the space the wedge occupies in a very straightforward manner, based on traditional XY-plane geometry.
Cylindrical to Cartesian Conversion
Converting from cylindrical coordinates to Cartesian coordinates involves a few straightforward formulas. This process is vital when coordinates are given in one system but need to be interpreted or graphed in another.
  • In cylindrical coordinates, a point is defined by \((r, \theta, z)\), where \(r\) is the distance from the \(z\)-axis, \(\theta\) is the angular coordinate, and \(z\) is the height.
  • To convert these into Cartesian coordinates, use the relations: \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\).
These conversions allow for easy manipulation and representation of complex 3D shapes, like the wedge in our exercise, across different coordinate systems.

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Most popular questions from this chapter

If \(f(0)=0,\) show that the curvature of the polar curve \(r=f(\theta)\) at \(\theta=0\) is given by \(\kappa=\frac{2}{\left|f^{\prime}(0)\right|}\).

To show that the surface in example 6.1 is the entire sphere \(x^{2}+y^{2}+z^{2}=4,\) start by finding the trace of the sphere in the plane \(z=k\) for \(-2 \leq k \leq 2 .\) If \(z=2 \cos v=k,\) determine as fully as possible the value of \(2 \sin v\) and then determine the trace in the plane \(z=k\) for \(x=2 \cos u \sin v, y=2 \sin u \sin v\) and \(z=2 \cos v .\) If the traces are the same, then the surfaces are the same.

If acceleration is parallel to position (a \(\| \mathbf{r}\) ), show that there is no torque. Explain this result in terms of the change in angular momentum. (Hint: If a l|r, would angular velocity or linear velocity be affected?)

In this exercise, we prove Kepler's third law. Recall that the area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(\pi a b .\) From exercise \(39,\) the rate at which area is swept out is given by \(\frac{d A}{d t}=\frac{1}{2}\|\mathbf{r} \times \mathbf{v}\|\) Conclude that the period of the orbit is \(T=\frac{\pi \bar{a} b}{\frac{1}{2}\|\mathbf{r} \times \mathbf{v}\|}\) and so, \(T^{2}=\frac{4 \pi^{2} a^{2} b^{2}}{\|\mathbf{r} \times \mathbf{v}\|^{2}} .\) Use (5.17) to show that the minimum value of \(r\) is \(r_{\min }=\frac{e d}{1+e}\) and that the maximum value of \(r\) is \(r_{\max }=\frac{e d}{1-e} .\) Explain why \(2 a=r_{\min }+r_{\max }\) and use this to show that \(a=\frac{e d}{1-e^{2}} .\) Given that \(1-e^{2}=\frac{b^{2}}{a^{2}},\) show that \(\frac{b^{2}}{a}=e d .\) From \(e=\frac{b}{G M}\) and \(d=\frac{c^{2}}{b},\) show that \(e d=\frac{\|\mathbf{r} \times \mathbf{v}\|^{2}}{G M} .\) It then follows that \(\frac{b^{2}}{a}=\frac{\|\mathbf{r} \times \mathbf{v}\|^{2}}{G M} .\) Finally, show that \(T^{2}=k a^{3},\) where the constant \(k=\frac{4 \pi^{2}}{G M}\) does not depend on the specific orbit of the planet.

Find parametric equations for the region that lies above \(z=\sqrt{x^{2}+y^{2}}\) and below \(x^{2}+y^{2}+z^{2}=4.\)
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