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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![### 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
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