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Eliminate the parameters to obtain an equation in rectangular coordinates, and describe the surface. $$ \begin{array}{l}{x=\sqrt{u} \cos v, y=\sqrt{u} \sin v, z=u \text { for } 0 \leq u \leq 4 \text { and }} \\ {0 \leq v<2 \pi}\end{array} $$

Short Answer

Expert verified
The surface is a paraboloid described by \( z = x^2 + y^2 \), constrained by \( 0 \leq z \leq 4 \).

Step by step solution

01

Identify the Relationships

Start with the parametric equations: \( x = \sqrt{u} \cos v \), \( y = \sqrt{u} \sin v \), and \( z = u \). We need to express these in terms of rectangular coordinates by eliminating the parameters \( u \) and \( v \).
02

Eliminate Parameter \( v \)

Use the identities of sine and cosine to eliminate \( v \). We have \( x = \sqrt{u} \cos v \) and \( y = \sqrt{u} \sin v \). Squaring both equations gives \( x^2 = u \cos^2 v \) and \( y^2 = u \sin^2 v \). Adding these results in \( x^2 + y^2 = u(\cos^2 v + \sin^2 v) = u \). This uses the Pythagorean identity \( \cos^2 v + \sin^2 v = 1 \).
03

Express \( z \) in terms of \( x \) and \( y \)

Since \( z = u \), and from Step 2 we found \( u = x^2 + y^2 \), substitute \( u \) in the \( z \) expression: \( z = x^2 + y^2 \).
04

Describe the Surface

The equation \( z = x^2 + y^2 \) represents a paraboloid opening upwards along the \( z \)-axis. The trace in any plane parallel to the \( xy \) plane (constant \( z \)) is a circle centered at the origin. Since \( 0 \leq u \leq 4 \), it restricts \( 0 \leq z \leq 4 \).

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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 powerful tool in mathematics, allowing us to describe complex curves and surfaces in a more intuitive way. These equations define the coordinates of points on a curve or surface as functions of one or more parameters. For example, in the given problem, the parametric equations are:
  • \( x = \sqrt{u} \cos v \)
  • \( y = \sqrt{u} \sin v \)
  • \( z = u \)
Here, \( u \) and \( v \) are parameters. Parametric equations are especially useful when modeling shapes in three dimensions, such as surfaces and volumes. They allow us to explore how the shape changes with different values of the parameters. This flexibility makes parametric equations a crucial tool in engineering, computer graphics, and physics.
Pythagorean identity
The Pythagorean identity is a fundamental concept in trigonometry and plays a crucial role in simplifying parametric equations. Specifically, it states that:\[\cos^2 v + \sin^2 v = 1\]This identity is derived from the Pythagorean Theorem and works for any angle \( v \). In the context of solving the given problem, the identity helps eliminate the parameter \( v \) from the parametric equations for \( x \) and \( y \):
  • By squaring \( x = \sqrt{u} \cos v \), we get \( x^2 = u \cos^2 v \).
  • Squaring \( y = \sqrt{u} \sin v \) gives \( y^2 = u \sin^2 v \).
  • Adding both equations, \( x^2 + y^2 = u(\cos^2 v + \sin^2 v) = u \).
This simplification using the Pythagorean identity is key to converting parametric equations into rectangular coordinates. It enables us to find the relationship between \( x \), \( y \), and \( z \) without the parameters, ultimately leading to the equation \( z = x^2 + y^2 \).
paraboloid surface
A paraboloid is a quadratic surface that can either open upward or downward, depending on its specific equation. In the exercise, we derive the equation of a paraboloid by transforming the parametric equations into rectangular coordinates, leading to:\[z = x^2 + y^2\]This indicates an upward-opening paraboloid with its vertex at the origin (0, 0, 0). Paraboloids have the unique geometric property where all cross-sections parallel to the xy-plane are circles, and sections parallel to the xz or yz planes are parabolas. For the derived paraboloid:
  • The base at \( z = 0 \) is a circle with radius \( \sqrt{u} \). However, because \( u \leq 4 \), the paraboloid is confined to \( 0 \leq z \leq 4 \).
  • At different heights \( z \), the radius of the cross-sectional circles increases as \( \sqrt{z} \).
Understanding paraboloids is essential in various fields, such as optics, where parabolic mirrors focus light, and in engineering structures influenced by quadratic shapes.

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Most popular questions from this chapter

Find the centroid of the solid. The solid in the first octant bounded by the surface \(z=x y\) and the planes \(z=0, x=2,\) and \(y=2\)

These exercises reference the Theorem of Pappus: If \(R\) is a bounded plane region and \(L\) is a line that lies in the plane of \(R\) such that \(R\) is entirely on one side of \(L,\) then the volume of the solid formed by revolving \(R\) about \(L\) is given by $$\text {volume}=(\text {area of } R) \cdot\left(\begin{array}{c}{\text {distance} \text {traveled}} \\ {\text {by the centroid}}\end{array}\right)$$ Perform the following steps to prove the Theorem of Pappus: (a) Introduce an xy-coordinate system so that \(L\) is along the \(y\) -axis and the region \(R\) is in the first quadrant. Partition \(R\) into rectangular subregions in the usual way and let \(R_{k}\) be a typical subregion of \(R\) with center \(\left(x_{k}^{*}, y_{k}^{*}\right)\) and area \(\Delta A_{k}=\Delta x_{k} \Delta y_{k} .\) Show that the volume generated by \(R_{k}\) as it revolves about \(L\) is $$2 \pi x_{k}^{*} \Delta x_{k} \Delta y_{k}=2 \pi x_{k}^{*} \Delta A_{k}$$ (b) Show that the volume generated by \(R\) as it revolves about \(L\) is $$ V=\iint_{R} 2 \pi x d A=2 \pi \cdot \bar{x} \cdot[\text { area of } R] $$

The transformation \(x=a u, y=b v(a>0, b>0)\) can be rewritten as \(x / a=u, y / b=v,\) and hence it maps the circular region $$ u^{2}+v^{2} \leq 1 $$ into the elliptical region $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 $$ In these exercises, perform the integration by transforming the elliptical region of integration into a circular region of integration and then evaluating the transformed integral in polar coordinates. \(\iint_{R} e^{-\left(x^{2}+4 y^{2}\right)} d A,\) where \(R\) is the region enclosed by the ellipse \(\left(x^{2} / 4\right)+y^{2}=1\)

Use the transformation \(u=x-2 y, v=2 x+y\) to find $$ \iint_{R} \frac{x-2 y}{2 x+y} d A $$ where \(R\) is the rectangular region enclosed by the lines \(x-2 y=1, x-2 y=4,2 x+y=1,2 x+y=3\)

Find the center of gravity of the solid bounded by the paraboloid \(z=1-x^{2}-y^{2}\) and the \(x y\) -plane, assuming the density to be \(\delta(x, y, z)=x^{2}+y^{2}+z^{2}\)

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