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...
Related questions
Question
Please explain how to solve, thank you!
![### Calculating the Area Under a Parabolic Curve
**Problem Statement:**
Find the area under the graph of \( x = 4 - y^2 \) on the interval \([0, 2]\).
**Graph Explanation:**
In the provided graph, a parabola opens to the left with its vertex at \( (4, 0) \). The function given is \( x = 4 - y^2 \). The shaded region, which we are asked to find the area of, extends from \( y = 0 \) to \( y = 2 \) along the y-axis.
**Options:**
- \(\frac{-16}{3}\)
- 0
- \(\frac{16}{3}\)
- 6
**Solution:**
To calculate the area under the curve \( x = 4 - y^2 \) from \( y = 0 \) to \( y = 2 \), we use the definite integral.
\[
\text{Area} = \int_{0}^{2} (4 - y^2) \, dy
\]
First, find the antiderivative:
\[
\int (4 - y^2) \, dy = 4y - \frac{y^3}{3}
\]
Now evaluate this antiderivative at the bounds \( y = 0 \) and \( y = 2 \):
\[
\left[ 4y - \frac{y^3}{3} \right]_{0}^{2} = \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(0) - \frac{0^3}{3} \right)
\]
\[
= \left( 8 - \frac{8}{3} \right) - \left( 0 - 0 \right)
\]
\[
= 8 - \frac{8}{3}
\]
\[
= \frac{24}{3} - \frac{8}{3}
\]
\[
= \frac{16}{3}
\]
**Conclusion:**
The area under the curve \( x = 4 - y^2 \) on the interval \([0, 2]\) is \(\frac{16}{3}\).
Thus, the correct answer is:
\[
\Large \boxed{\frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F883933e7-6864-477d-8bb2-712659e12b95%2F33549a29-a402-4e3c-90f7-985f40ada29b%2F2lsr9bb_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Area Under a Parabolic Curve
**Problem Statement:**
Find the area under the graph of \( x = 4 - y^2 \) on the interval \([0, 2]\).
**Graph Explanation:**
In the provided graph, a parabola opens to the left with its vertex at \( (4, 0) \). The function given is \( x = 4 - y^2 \). The shaded region, which we are asked to find the area of, extends from \( y = 0 \) to \( y = 2 \) along the y-axis.
**Options:**
- \(\frac{-16}{3}\)
- 0
- \(\frac{16}{3}\)
- 6
**Solution:**
To calculate the area under the curve \( x = 4 - y^2 \) from \( y = 0 \) to \( y = 2 \), we use the definite integral.
\[
\text{Area} = \int_{0}^{2} (4 - y^2) \, dy
\]
First, find the antiderivative:
\[
\int (4 - y^2) \, dy = 4y - \frac{y^3}{3}
\]
Now evaluate this antiderivative at the bounds \( y = 0 \) and \( y = 2 \):
\[
\left[ 4y - \frac{y^3}{3} \right]_{0}^{2} = \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(0) - \frac{0^3}{3} \right)
\]
\[
= \left( 8 - \frac{8}{3} \right) - \left( 0 - 0 \right)
\]
\[
= 8 - \frac{8}{3}
\]
\[
= \frac{24}{3} - \frac{8}{3}
\]
\[
= \frac{16}{3}
\]
**Conclusion:**
The area under the curve \( x = 4 - y^2 \) on the interval \([0, 2]\) is \(\frac{16}{3}\).
Thus, the correct answer is:
\[
\Large \boxed{\frac
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images

Recommended textbooks for you

Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning

Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON

Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON

Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning

Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON

Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON

Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman


Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning