The equation y = Va? – x² ,-a < x s a, is the upper half of the circle centered at the origin with radius of a. Rotate this curve about the x-axis generate the sphere centered at the origin with radius equals to a. Use the disk method to show that: Volume of a sphere with radius of a is: V = Ta³ . Must work the integral out by hand and show all steps.
The equation y = Va? – x² ,-a < x s a, is the upper half of the circle centered at the origin with radius of a. Rotate this curve about the x-axis generate the sphere centered at the origin with radius equals to a. Use the disk method to show that: Volume of a sphere with radius of a is: V = Ta³ . Must work the integral out by hand and show all steps.
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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![### How to Derive the Volume of a Sphere Using the Disk Method
#### Equation and Setup
The equation \( y = \sqrt{a^2 - x^2} \), where \(-a \leq x \leq a\), represents the upper half of a circle centered at the origin with radius \(a\). By rotating this curve about the x-axis, a sphere centered at the origin with a radius of \(a\) can be generated. To find the volume of this sphere, we use the disk method. The disk method involves integrating the area of circular disks along the axis of rotation.
#### Volume Calculation Using the Disk Method
The volume \(V\) of a sphere with radius \(a\) can be derived using the integral:
\[ V = \int_{-a}^{a} \pi y^2 \, dx \]
Given \( y = \sqrt{a^2 - x^2} \), substitute \( y^2 \) with \( a^2 - x^2 \):
\[ V = \int_{-a}^{a} \pi (a^2 - x^2) \, dx \]
This simplifies to:
\[ V = \pi \int_{-a}^{a} (a^2 - x^2) \, dx \]
Breaking the integral into two parts:
\[ V = \pi \left[ \int_{-a}^{a} a^2 \, dx - \int_{-a}^{a} x^2 \, dx \right] \]
Since \(a^2\) is a constant, the first integral becomes:
\[ \int_{-a}^{a} a^2 \, dx = a^2 \int_{-a}^{a} \, dx = a^2 \left[ x \right]_{-a}^{a} = a^2 \left( a - (-a) \right) = 2a^3 \]
The second integral, \(\int_{-a}^{a} x^2 \, dx\), can be found using the power rule:
\[ \int_{-a}^{a} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-a}^{a} = \frac{a^3}{3} - \left( - \frac{a^3}{3} \right) = \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99269e07-9031-4b41-ba3f-5f1d3d15b266%2F3d079e17-b9ab-467a-b821-43ad0dfaf2af%2Fh28weqj.png&w=3840&q=75)
Transcribed Image Text:### How to Derive the Volume of a Sphere Using the Disk Method
#### Equation and Setup
The equation \( y = \sqrt{a^2 - x^2} \), where \(-a \leq x \leq a\), represents the upper half of a circle centered at the origin with radius \(a\). By rotating this curve about the x-axis, a sphere centered at the origin with a radius of \(a\) can be generated. To find the volume of this sphere, we use the disk method. The disk method involves integrating the area of circular disks along the axis of rotation.
#### Volume Calculation Using the Disk Method
The volume \(V\) of a sphere with radius \(a\) can be derived using the integral:
\[ V = \int_{-a}^{a} \pi y^2 \, dx \]
Given \( y = \sqrt{a^2 - x^2} \), substitute \( y^2 \) with \( a^2 - x^2 \):
\[ V = \int_{-a}^{a} \pi (a^2 - x^2) \, dx \]
This simplifies to:
\[ V = \pi \int_{-a}^{a} (a^2 - x^2) \, dx \]
Breaking the integral into two parts:
\[ V = \pi \left[ \int_{-a}^{a} a^2 \, dx - \int_{-a}^{a} x^2 \, dx \right] \]
Since \(a^2\) is a constant, the first integral becomes:
\[ \int_{-a}^{a} a^2 \, dx = a^2 \int_{-a}^{a} \, dx = a^2 \left[ x \right]_{-a}^{a} = a^2 \left( a - (-a) \right) = 2a^3 \]
The second integral, \(\int_{-a}^{a} x^2 \, dx\), can be found using the power rule:
\[ \int_{-a}^{a} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-a}^{a} = \frac{a^3}{3} - \left( - \frac{a^3}{3} \right) = \frac
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