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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Concept explainers
Cylinders
A cylinder is a three-dimensional solid shape with two parallel and congruent circular bases, joined by a curved surface at a fixed distance. A cylinder has an infinite curvilinear surface.
Cones
A cone is a three-dimensional solid shape having a flat base and a pointed edge at the top. The flat base of the cone tapers smoothly to form the pointed edge known as the apex. The flat base of the cone can either be circular or elliptical. A cone is drawn by joining the apex to all points on the base, using segments, lines, or half-lines, provided that the apex and the base both are in different planes.
Question
![**Problem 3:**
A circular cone has a height of 10 units and a base radius of 4 units. Find the average area of a circular cross-section of the cone.
---
To solve this problem, consider the cone's cross-sections parallel to the base. If you take a horizontal slice of the cone at a height \( h \), the radius \( r \) of the slice is proportional to its height from the vertex.
### Explanation:
- The radius \( r \) at any height \( x \) from the vertex is given by the formula:
\[
r = \frac{4}{10} \times (10-x) = 2(1-x/5)
\]
- The area \( A \) of a cross-section is \( \pi r^2 \):
\[
A = \pi \left(2\left(1-\frac{x}{5}\right)\right)^2 = 4\pi \left(1-\frac{x}{5}\right)^2
\]
### Average Area:
To find the average area of these cross-sections from \( x = 0 \) to \( x = 10 \) (top to bottom of the cone), integrate the area function and then divide by the height:
1. Integrate \( A(x) \) from 0 to 10:
\[
\int_{0}^{10} 4\pi \left(1-\frac{x}{5}\right)^2 \, dx
\]
2. Find the average:
\[
\frac{1}{10} \int_{0}^{10} 4\pi \left(1-\frac{x}{5}\right)^2 \, dx
\]
By solving the integral and dividing by 10, you get the average cross-sectional area.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e9683aa-2143-4756-b701-dd7fd0ed0552%2F51d6385b-b7ea-45a7-8476-70ae5474b67d%2Flosu4ji_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 3:**
A circular cone has a height of 10 units and a base radius of 4 units. Find the average area of a circular cross-section of the cone.
---
To solve this problem, consider the cone's cross-sections parallel to the base. If you take a horizontal slice of the cone at a height \( h \), the radius \( r \) of the slice is proportional to its height from the vertex.
### Explanation:
- The radius \( r \) at any height \( x \) from the vertex is given by the formula:
\[
r = \frac{4}{10} \times (10-x) = 2(1-x/5)
\]
- The area \( A \) of a cross-section is \( \pi r^2 \):
\[
A = \pi \left(2\left(1-\frac{x}{5}\right)\right)^2 = 4\pi \left(1-\frac{x}{5}\right)^2
\]
### Average Area:
To find the average area of these cross-sections from \( x = 0 \) to \( x = 10 \) (top to bottom of the cone), integrate the area function and then divide by the height:
1. Integrate \( A(x) \) from 0 to 10:
\[
\int_{0}^{10} 4\pi \left(1-\frac{x}{5}\right)^2 \, dx
\]
2. Find the average:
\[
\frac{1}{10} \int_{0}^{10} 4\pi \left(1-\frac{x}{5}\right)^2 \, dx
\]
By solving the integral and dividing by 10, you get the average cross-sectional area.
---
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