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Find a parametric representation of the surface in terms of the parameters \(r\) and \(\theta,\) where \((r, \theta, z)\) are the cylindrical coordinates of a point on the surface. $$ z=e^{-\left(x^{2}+y^{2}\right)} $$

Short Answer

Expert verified
The parametric representation is \(\mathbf{r}(r, \theta) = (r \cos(\theta), r \sin(\theta), e^{-r^2})\).

Step by step solution

01

Express Cartesian coordinates in terms of cylindrical coordinates

In cylindrical coordinates, the relationships are given by: - \( x = r \cos(\theta) \) - \( y = r \sin(\theta) \). Here, \( r \) is the radius (distance from the z-axis), and \( \theta \) is the angle in the xy-plane from the positive x-axis.
02

Substitute cylindrical coordinates into the equation for z

The given surface equation is \( z = e^{-(x^2 + y^2)} \). Use the expressions for \( x \) and \( y \) in terms of \( r \) and \( \theta \):\[ x^2 + y^2 = (r \cos(\theta))^2 + (r \sin(\theta))^2 = r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2(\cos^2(\theta) + \sin^2(\theta)) = r^2 \]Thus, the equation becomes \( z = e^{-r^2} \).
03

Formulate the parametric equations of the surface

A parametric representation of the surface can be written as a vector equation using the parameters \( r \) and \( \theta \). The surface is given by:\[ \mathbf{r}(r, \theta) = (r \cos(\theta), r \sin(\theta), e^{-r^2}) \] This represents the original surface in terms of cylindrical coordinates.

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Key Concepts

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

Parametric Representation: Expressing Surfaces with Parameters
A parametric representation of a surface involves describing it using parameters, which simplify the relationship between the surface's coordinates. By choosing specific parameters, like \( r \) and \( \theta \) in cylindrical coordinates, you can easily navigate and manipulate complex surfaces. This representation breaks down complex geometrical forms into simpler, more manageable expressions.

In the context of our original problem, we're exploring a surface given by the equation \( z = e^{-(x^2 + y^2)} \). To convert this into a parametric form using the parameters \( r \) (radius) and \( \theta \) (angle), we first express \( x \) and \( y \) in terms of \( r \) and \( \theta \):
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
By substituting these expressions into the original surface equation, we simplify \( z \) to \( z = e^{-r^2} \).
The resulting parametric representation, \( \mathbf{r}(r, \theta) = (r \cos(\theta), r \sin(\theta), e^{-r^2}) \), encapsulates the entire surface with the use of two parameters, making the computations much simpler.
Understanding the Surface Equation: The Role of Exponents
Surface equations often describe complex three-dimensional shapes. In this exercise, the surface is defined by the equation \( z = e^{-(x^2 + y^2)} \). This equation uses the exponential function, which is crucial in many mathematical applications due to its unique properties.

The expression \( x^2 + y^2 \) in the exponent represents the squared distance from the origin on the xy-plane, which simplifies to \( r^2 \) when translated into cylindrical coordinates:
  • The term \( r^2 \) naturally appears because of the faint role of the Pythagorean theorem in recognizing radial distance.
  • As you plot \( z = e^{-r^2} \), you'll notice that the surface forms a sort of smooth mound or "bell curve" around the origin, tapering off rapidly as \( r \) increases.
This specific surface is an example of a Gaussian function, commonly appearing in probability and statistics as well as physics.
Coordinate Transformation in Cylindrical Coordinates
Coordinate transformation allows us to reframe problems using different coordinate systems that might better fit the symmetry of the problem. In this case, switching to cylindrical coordinates from Cartesian coordinates simplifies navigating the given surface equation.

Cylindrical coordinates are incredibly useful when dealing with rotational or radial symmetry, as they inherently align with these types of physical and geometric configurations. The transformation from Cartesian to cylindrical involves the following relationships:
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
  • \( z = z \) (no change, as both systems share this axis)
Utilizing these transformations, the equation \( z = e^{-(x^2 + y^2)} \) simplifies beautifully to \( z = e^{-r^2} \). This transformation dramatically streamlines calculations, offering a more intuitive feel for the shape and form of the surface in question. It emphasizes the radial symmetry that wasn't as obvious in the Cartesian coordinates.

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Most popular questions from this chapter

The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region \(R\) of the \(x y\) -plane, and if its density function \(\delta(x, y)\) is continuous on \(R,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iint_{R} y^{2} \delta(x, y) d A, \quad I_{y}=\iint_{R} x^{2} \delta(x, y) d A \\ I_{z} &=\iint_{R}\left(x^{2}+y^{2}\right) \delta(x, y) d A \end{aligned}$$ Use these definitions in Exercises 45 and 46 Consider the circular lamina that occupies the region described by the inequalities \(0 \leq x^{2}+y^{2} \leq a^{2} .\) Assuming that the lamina has constant density \(\delta,\) show that $$ I_{x}=I_{y}=\frac{\delta \pi a^{4}}{4}, \quad I_{z}=\frac{\delta \pi a^{4}}{2} $$

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Find the center of gravity of the cube that is determined by the inequalities \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\) if (a) the density is proportional to the square of the distance to the origin; (b) the density is proportional to the sum of the distances to the faces that lie in the coordinate planes.

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