Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem Statement: Surface Area of a Cone**
Objective: Find the surface area of the cone defined by the equation \( z = \sqrt{x^2 + y^2} \) within the range \( 0 \leq z \leq 1 \). Achieve this by establishing a proper parametrization first.
**Explanation:**
This problem involves calculating the surface area of a cone using calculus concepts. The given cone's equation, \( z = \sqrt{x^2 + y^2} \), suggests that z describes the height of the cone as a function of \( x \) and \( y \). The constraint \( 0 \leq z \leq 1 \) limits the height of the cone to this range, essentially defining a portion of the cone with a capped height.
To solve this problem, it is necessary to first find a parametrization that can express the surface of this cone in a more manageable form. Parametrization involves expressing the coordinates \( x \), \( y \), and \( z \) in terms of two parameters, typically denoted as \( u \) and \( v \).
**Solution Steps:**
1. **Parametrize the Cone**:
- Choose a parametrization that expresses \( x \), \( y \), and \( z \) in terms of parameters \( u \) and \( v \). A common choice for a cone is to use cylindrical coordinates. For instance:
- Let \( x = u \cos(v) \)
- Let \( y = u \sin(v) \)
- Let \( z = u \)
- Here, \( u \) represents the radius on the base of the cone (ranges from 0 to 1 due to the constraint on \( z \)), and \( v \) is the angular component (ranges from 0 to \( 2\pi \)).
2. **Calculate the Surface Area**:
- Once the surface is parametrized, use the surface area formula for parametric surfaces:
\[
A = \int \int \left| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right| \, du \, dv
\]
- Here, \( \mathbf{r}(u, v) = (u \cos(v), u \sin(v), u) \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83a2909f-e771-46af-ac38-b46aeee99cc8%2Fc04687bb-67f4-4c7d-a8d2-02a013f26442%2F99sjl0m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Surface Area of a Cone**
Objective: Find the surface area of the cone defined by the equation \( z = \sqrt{x^2 + y^2} \) within the range \( 0 \leq z \leq 1 \). Achieve this by establishing a proper parametrization first.
**Explanation:**
This problem involves calculating the surface area of a cone using calculus concepts. The given cone's equation, \( z = \sqrt{x^2 + y^2} \), suggests that z describes the height of the cone as a function of \( x \) and \( y \). The constraint \( 0 \leq z \leq 1 \) limits the height of the cone to this range, essentially defining a portion of the cone with a capped height.
To solve this problem, it is necessary to first find a parametrization that can express the surface of this cone in a more manageable form. Parametrization involves expressing the coordinates \( x \), \( y \), and \( z \) in terms of two parameters, typically denoted as \( u \) and \( v \).
**Solution Steps:**
1. **Parametrize the Cone**:
- Choose a parametrization that expresses \( x \), \( y \), and \( z \) in terms of parameters \( u \) and \( v \). A common choice for a cone is to use cylindrical coordinates. For instance:
- Let \( x = u \cos(v) \)
- Let \( y = u \sin(v) \)
- Let \( z = u \)
- Here, \( u \) represents the radius on the base of the cone (ranges from 0 to 1 due to the constraint on \( z \)), and \( v \) is the angular component (ranges from 0 to \( 2\pi \)).
2. **Calculate the Surface Area**:
- Once the surface is parametrized, use the surface area formula for parametric surfaces:
\[
A = \int \int \left| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right| \, du \, dv
\]
- Here, \( \mathbf{r}(u, v) = (u \cos(v), u \sin(v), u) \)
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