21. Find the volume of the solid bounded above by the surface z = ryery and below by the rectangle R: 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1. (Set up and evaluate a double integral).
21. Find the volume of the solid bounded above by the surface z = ryery and below by the rectangle R: 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1. (Set up and evaluate a double integral).
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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Can you please explain the solution in the attached image to this problem? I don't know where the bounds came from in the second half of the answer.
Transcribed Image Text:21. Find the volume of the solid bounded above by the surface z = Tyety and below by the
rectangle R: 0≤x≤ 2 and 0 ≤ y ≤ 1. (Set up and evaluate a double integral).
![= £²6²² xyey² dydx. For the inner integral (with respect to y), we will use u-sub:
u = xy², du = u'dy = 2xydy ⇒ dy = 2du, the y-limits become u(0) = 0 and u(1) = x and
the volume integral becomes:
21. Volume V =
"I
I
1
V
=
- [² [²* zye" Lydu]dz - f²f²c²/2 du]dx = ª [cª/2]* dx = √² [c²/2 - 1/2] dz
²
[
dx =
2xy
= [e² /2 − x/2]² = (e² /2 − 1) − (1/2 − 0) = e²/2 − 3/2.
-
0
13³1² [x²=*
√3x
r3/2
-3/2
- [³ ² xy² √²²" dx = ["¹" [2(9-x^²) = 3x³) dr
0
√√3x
0
2xy dyda = ay
3/2
= 1³th² (9x - - 4x³) dx
= [(9/2)x² − x4]3/² =
=
0
= (9/4)(9/2-9/4)= 81/16.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6389447-1237-4af0-b5c6-eb1260425b55%2Fa3db2e54-cb73-4ab2-9ae1-87a5821600b0%2Fedpf3db_processed.png&w=3840&q=75)
Transcribed Image Text:= £²6²² xyey² dydx. For the inner integral (with respect to y), we will use u-sub:
u = xy², du = u'dy = 2xydy ⇒ dy = 2du, the y-limits become u(0) = 0 and u(1) = x and
the volume integral becomes:
21. Volume V =
"I
I
1
V
=
- [² [²* zye" Lydu]dz - f²f²c²/2 du]dx = ª [cª/2]* dx = √² [c²/2 - 1/2] dz
²
[
dx =
2xy
= [e² /2 − x/2]² = (e² /2 − 1) − (1/2 − 0) = e²/2 − 3/2.
-
0
13³1² [x²=*
√3x
r3/2
-3/2
- [³ ² xy² √²²" dx = ["¹" [2(9-x^²) = 3x³) dr
0
√√3x
0
2xy dyda = ay
3/2
= 1³th² (9x - - 4x³) dx
= [(9/2)x² − x4]3/² =
=
0
= (9/4)(9/2-9/4)= 81/16.
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