I don't understand why the double integral of z dA over region D turns into [(64-4r^2)^(1/2)-(-(64-4r^2)^(1/2)]rdrdtheta, instead of just (64-4r^2)^(1/2) rdrdtheta since R {(r,theta) | 0 <= r <= 2; 0 <= theta <+ 2pi}.
I don't understand why the double integral of z dA over region D turns into [(64-4r^2)^(1/2)-(-(64-4r^2)^(1/2)]rdrdtheta, instead of just (64-4r^2)^(1/2) rdrdtheta since R {(r,theta) | 0 <= r <= 2; 0 <= theta <+ 2pi}.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I don't understand why the double
![← Essential Calculus: E... 2nd Edition
Given:
The region D lies inside the cylinders x² + y² = 4 and the ellipsoid 4x² + 4y² + z² = 64.
Formula used:
If f is a polar rectangle R given by 0 < a < r ≤ b, a ≤ 0 ≤ B, where 0 ≤B-a ≤ 2π, then,
B b
[[ ƒ (z,y) dA = [[ƒ (r cos 0, r sin 0) rdrdė (1)
R
a a
If g (x) is the function of x and h (y) is the function of y then,
b d
[ [ 9 (2) h (y) dydx = [ 9(x) dx | h (y) dy (2)
ac
Calculation:
Obtain the value for z from the ellipsoid.
4x² + 4y² + x² = 64
2² 64-4x² - 4y²
=
z =
±√64-4x² - 4y²
z = ±√√/64-4 (x² + y²)
And, the value of r varies from 0 to 2 and the value of varies from 0 to 2πT.
Substitute x = r cos 0 and y = r sin in the equation (1) and obtain the required volume.
2T 2
|| -
zdA=
D
00
[√₁²
√64 – 4r² − (−√64 – 4r²)] (r) drdė
2π 2
= 2 (2) ] [r√/16 - 7² drdo
00
2 2
1 [ [TVIE
0
Chapter 12.3, Problem 19E
= 4
r√16-r²drde](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93013f87-eae9-408f-8367-c564e6d22e0c%2F25686ff9-225a-4910-a873-013dbdc935c4%2F7jdhooq_processed.png&w=3840&q=75)
Transcribed Image Text:← Essential Calculus: E... 2nd Edition
Given:
The region D lies inside the cylinders x² + y² = 4 and the ellipsoid 4x² + 4y² + z² = 64.
Formula used:
If f is a polar rectangle R given by 0 < a < r ≤ b, a ≤ 0 ≤ B, where 0 ≤B-a ≤ 2π, then,
B b
[[ ƒ (z,y) dA = [[ƒ (r cos 0, r sin 0) rdrdė (1)
R
a a
If g (x) is the function of x and h (y) is the function of y then,
b d
[ [ 9 (2) h (y) dydx = [ 9(x) dx | h (y) dy (2)
ac
Calculation:
Obtain the value for z from the ellipsoid.
4x² + 4y² + x² = 64
2² 64-4x² - 4y²
=
z =
±√64-4x² - 4y²
z = ±√√/64-4 (x² + y²)
And, the value of r varies from 0 to 2 and the value of varies from 0 to 2πT.
Substitute x = r cos 0 and y = r sin in the equation (1) and obtain the required volume.
2T 2
|| -
zdA=
D
00
[√₁²
√64 – 4r² − (−√64 – 4r²)] (r) drdė
2π 2
= 2 (2) ] [r√/16 - 7² drdo
00
2 2
1 [ [TVIE
0
Chapter 12.3, Problem 19E
= 4
r√16-r²drde
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