In this question we use two ways to calculate an integral and in doing so, verify the Generalized Stokes's Theorem. Integrate over the oriented boundary of the manifold Min R³ consisting of those points x satisfying Part 1. First we compute = ydx / dz, ²+590≤3. 4. Since M is a cylinder in R³, it's boundary consists of three parts, and we integrate each one separately. вм For the side surface S₁ of the cylinder, we parametrize it using = 3 cos, y = 3 sinu and z = z. So P₁ = (C ◉◉ [+] (2) [L <= dudz= For the disc S₂ at the top of the cylinder, we parametrize it using dudz= P₁ = Pr оо Op Op 4= $ B₂ Dr Ou Finally for the disc Sy at the bottom of the cylinder, we parametrize it using drdu= P₁ = (0 = др др drdu= Dr' Ou +2 Part 2. Now let's compute q= q= 4== -L- Now: M M оо do. To begin we parametrize M: O O O ㅁㅁㅁ dsdud:= drdu= p = (3 cosu, 3 sinu, z), u € [0,2], = = [0,3]. p = (rcosu,rsinu, 3), [0, 2π], r € [0,3]. p= 14 p=(rcosu, rsinu, 0), [0,2], † € [0,3], q= (scosu, 8sinu, ), 8€ [0,3], µ = [0,2], z ¤ [0,3].

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.3: Volume And Average Value
Problem 17E
Question
In this question we use two ways to calculate an integral and in doing so, verify the Generalized Stokes's Theorem.
Integrate
over the oriented boundary of the manifold Min R³ consisting of those points x satisfying
Part 1. First we compute
= ydx / dz,
²+590≤3.
4. Since M is a cylinder in R³, it's boundary consists of three parts, and we integrate each one separately.
вм
For the side surface S₁ of the cylinder, we parametrize it using = 3 cos, y = 3 sinu and z = z. So
P₁ =
(C
◉◉
[+] (2) [L
<=
dudz=
For the disc S₂ at the top of the cylinder, we parametrize it using
dudz=
P₁ =
Pr
оо
Op Op
4=
$
B₂
Dr Ou
Finally for the disc Sy at the bottom of the cylinder, we parametrize it using
drdu=
P₁ =
(0
=
др др
drdu=
Dr' Ou
+2
Part 2. Now let's compute
q=
q=
4==
-L-
Now:
M
M
оо
do. To begin we parametrize M:
O O O
ㅁㅁㅁ
dsdud:=
drdu=
p = (3 cosu, 3 sinu, z), u € [0,2], = = [0,3].
p = (rcosu,rsinu, 3), [0, 2π], r € [0,3].
p=
14
p=(rcosu, rsinu, 0), [0,2], † € [0,3],
q= (scosu, 8sinu, ), 8€ [0,3], µ = [0,2], z ¤ [0,3].
Transcribed Image Text:In this question we use two ways to calculate an integral and in doing so, verify the Generalized Stokes's Theorem. Integrate over the oriented boundary of the manifold Min R³ consisting of those points x satisfying Part 1. First we compute = ydx / dz, ²+590≤3. 4. Since M is a cylinder in R³, it's boundary consists of three parts, and we integrate each one separately. вм For the side surface S₁ of the cylinder, we parametrize it using = 3 cos, y = 3 sinu and z = z. So P₁ = (C ◉◉ [+] (2) [L <= dudz= For the disc S₂ at the top of the cylinder, we parametrize it using dudz= P₁ = Pr оо Op Op 4= $ B₂ Dr Ou Finally for the disc Sy at the bottom of the cylinder, we parametrize it using drdu= P₁ = (0 = др др drdu= Dr' Ou +2 Part 2. Now let's compute q= q= 4== -L- Now: M M оо do. To begin we parametrize M: O O O ㅁㅁㅁ dsdud:= drdu= p = (3 cosu, 3 sinu, z), u € [0,2], = = [0,3]. p = (rcosu,rsinu, 3), [0, 2π], r € [0,3]. p= 14 p=(rcosu, rsinu, 0), [0,2], † € [0,3], q= (scosu, 8sinu, ), 8€ [0,3], µ = [0,2], z ¤ [0,3].
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