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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This is a solved problem already, but please can you explain to me how we came out with the (highlighted yellow) part !!
Thank you ! 
![### Triple Integral Calculation
The expression of the triple integral over volume \( V \) is given by:
\[
\iiint_V z \, dV = \int_0^a \int_0^{b(1-x/a)} \int_0^{c(1-x/a-y/b)} z \, dz \, dy \, dx
\]
#### Inner Integral
The inner integral is:
\[
\int_0^{c(1-x/a-y/b)} z \, dz = \left[\frac{1}{2} z^2 \right]_{z=0}^{z=c(1-x/a-y/b)} = \frac{1}{2} c^2 \left(1 - \frac{x}{a} - \frac{y}{b} \right)^2
\]
#### Middle Integral
The middle integral becomes:
\[
\frac{1}{2} c^2 \int_0^{b(1-x/a)} \left(1 - \frac{x}{a} - \frac{y}{b} \right)^2 \, dy = \frac{1}{2} c^2 \left[ -\frac{b}{3} \left(1 - \frac{x}{a} - \frac{y}{b} \right)^3 \right]_{y=0}^{y=b(1-x/a)}
\]
which simplifies to:
\[
= \frac{1}{6} bc^2 \left(1 - \frac{x}{a} \right)^3
\]
#### Outer Integral
The outer integral is then:
\[
\frac{1}{6} bc^2 \int_0^a \left(1 - \frac{x}{a} \right)^3 \, dx = \frac{1}{6} bc^2 \left[ -\frac{a}{4} \left(1 - \frac{x}{a} \right)^4 \right]_0^a
\]
which evaluates to:
\[
= \frac{1}{24} abc^2
\]
Thus, the final result of the triple integration is:
\[
\boxed{\frac{1}{24} abc^2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa41da8ad-0c60-4ee0-8453-d772480812a3%2F4ed1e5b9-9b0c-43f3-b161-e7797b3d7614%2Fkw8390j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Triple Integral Calculation
The expression of the triple integral over volume \( V \) is given by:
\[
\iiint_V z \, dV = \int_0^a \int_0^{b(1-x/a)} \int_0^{c(1-x/a-y/b)} z \, dz \, dy \, dx
\]
#### Inner Integral
The inner integral is:
\[
\int_0^{c(1-x/a-y/b)} z \, dz = \left[\frac{1}{2} z^2 \right]_{z=0}^{z=c(1-x/a-y/b)} = \frac{1}{2} c^2 \left(1 - \frac{x}{a} - \frac{y}{b} \right)^2
\]
#### Middle Integral
The middle integral becomes:
\[
\frac{1}{2} c^2 \int_0^{b(1-x/a)} \left(1 - \frac{x}{a} - \frac{y}{b} \right)^2 \, dy = \frac{1}{2} c^2 \left[ -\frac{b}{3} \left(1 - \frac{x}{a} - \frac{y}{b} \right)^3 \right]_{y=0}^{y=b(1-x/a)}
\]
which simplifies to:
\[
= \frac{1}{6} bc^2 \left(1 - \frac{x}{a} \right)^3
\]
#### Outer Integral
The outer integral is then:
\[
\frac{1}{6} bc^2 \int_0^a \left(1 - \frac{x}{a} \right)^3 \, dx = \frac{1}{6} bc^2 \left[ -\frac{a}{4} \left(1 - \frac{x}{a} \right)^4 \right]_0^a
\]
which evaluates to:
\[
= \frac{1}{24} abc^2
\]
Thus, the final result of the triple integration is:
\[
\boxed{\frac{1}{24} abc^2}
\]
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