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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![### Solving a Triple Integral Equation for \( c \)
In this example, we are given a triple integral equation that we need to solve for \( c \). The equation is as follows:
\[ \int_{0}^{1} \int_{0}^{1 - x^2} \int_{c}^{1 - x^2 - y} dz \, dy \, dx = \frac{4}{15} \]
This equation involves integrating over a three-dimensional region where the bounds of integration are dependent on the variables \( x \) and \( y \).
Here's a breakdown of the equation:
- The outermost integral is with respect to \( x \), with limits from 0 to 1.
- The middle integral is with respect to \( y \), with limits from 0 to \( 1 - x^2 \).
- The innermost integral is with respect to \( z \), with limits from \( c \) to \( 1 - x^2 - y \).
The right-hand side of the equation is given by the fraction \( \frac{4}{15} \).
To solve for \( c \):
1. Compute the innermost integral with respect to \( z \).
2. Substitute the result into the middle integral and compute with respect to \( y \).
3. Finally, substitute this result into the outermost integral and compute with respect to \( x \).
4. Set the resulting expression equal to \( \frac{4}{15} \) and solve for \( c \).
Note: The detailed steps of solving this integral algebraically are complex and usually covered in advanced calculus or multivariable calculus courses where techniques such as Fubini's Theorem may be used.
This type of problem is common in courses focusing on multivariable calculus and integral calculus.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d576798-e517-451f-b302-9c2c35ed41b2%2F7140c7f3-e150-4711-92de-bffa47067e47%2Fjbamhn_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving a Triple Integral Equation for \( c \)
In this example, we are given a triple integral equation that we need to solve for \( c \). The equation is as follows:
\[ \int_{0}^{1} \int_{0}^{1 - x^2} \int_{c}^{1 - x^2 - y} dz \, dy \, dx = \frac{4}{15} \]
This equation involves integrating over a three-dimensional region where the bounds of integration are dependent on the variables \( x \) and \( y \).
Here's a breakdown of the equation:
- The outermost integral is with respect to \( x \), with limits from 0 to 1.
- The middle integral is with respect to \( y \), with limits from 0 to \( 1 - x^2 \).
- The innermost integral is with respect to \( z \), with limits from \( c \) to \( 1 - x^2 - y \).
The right-hand side of the equation is given by the fraction \( \frac{4}{15} \).
To solve for \( c \):
1. Compute the innermost integral with respect to \( z \).
2. Substitute the result into the middle integral and compute with respect to \( y \).
3. Finally, substitute this result into the outermost integral and compute with respect to \( x \).
4. Set the resulting expression equal to \( \frac{4}{15} \) and solve for \( c \).
Note: The detailed steps of solving this integral algebraically are complex and usually covered in advanced calculus or multivariable calculus courses where techniques such as Fubini's Theorem may be used.
This type of problem is common in courses focusing on multivariable calculus and integral calculus.
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