I don't understand why the -xy transform to -r^2*cos^2(theta)*sin(theta). Since I think x=rcos(theta), y=rsin(theta), -xy should be equal to -r^2*cos(theta)*sin(theta). where does cos squared come from？
I don't understand why the -xy transform to -r^2*cos^2(theta)*sin(theta). Since I think x=rcos(theta), y=rsin(theta), -xy should be equal to -r^2*cos(theta)*sin(theta). where does cos squared come from？
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 -xy transform to -r^2*cos^2(theta)*sin(theta). Since I think x=rcos(theta), y=rsin(theta), -xy should be equal to -r^2*cos(theta)*sin(theta). where does cos squared come from？
![Recall that by the divergence theorem
2- ds = [₂²
where D is the solid bounded by the surface S.
Using the definition of divergence we get
F.
V.F= 2xy - 2xy - xy = -xy. (2)
The hyperboloid has only one sheet and it is a surface of revolution around z-axis. Note
that
z = [-2, 2]
and the corresponding z-slice is a half disc x² + y² = 2² +1, y > 0. On these slices
we can use polar coordinates to evaluate the double integral. That transforms the area
{x² + y² ≤ z²+1, y>0} into {0 < r ≤ √z² +1,0 ≤ 0 ≤ T}. 4
That said, we get
]][₁·E = £²₂ rydz dy) dz
V.F
dx
=
V.FdV
=
{x²+y²<z²+1,y>0}
√z²+1 Cπ
- L. (C S
T
=
L (²² 0 ² dr) d² ²006 ²018?
dz
3
10
0
/2²+1
2
3
- 3 L₁ (1² +³ dr) d=
=
dz
0
r² cos² sin 0r de dr dz
dr) dz
L² (²² + 1)² dz
6
1 ² (2²¹ +22² + 1) dz =
206
45
(4)
(4)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F73001e82-7a92-4cbf-ae31-e9298bb78d6a%2F554614ba-e1c4-4a4a-b0ba-32fbaf1588ba%2F6oelanf_processed.png&w=3840&q=75)
Transcribed Image Text:Recall that by the divergence theorem
2- ds = [₂²
where D is the solid bounded by the surface S.
Using the definition of divergence we get
F.
V.F= 2xy - 2xy - xy = -xy. (2)
The hyperboloid has only one sheet and it is a surface of revolution around z-axis. Note
that
z = [-2, 2]
and the corresponding z-slice is a half disc x² + y² = 2² +1, y > 0. On these slices
we can use polar coordinates to evaluate the double integral. That transforms the area
{x² + y² ≤ z²+1, y>0} into {0 < r ≤ √z² +1,0 ≤ 0 ≤ T}. 4
That said, we get
]][₁·E = £²₂ rydz dy) dz
V.F
dx
=
V.FdV
=
{x²+y²<z²+1,y>0}
√z²+1 Cπ
- L. (C S
T
=
L (²² 0 ² dr) d² ²006 ²018?
dz
3
10
0
/2²+1
2
3
- 3 L₁ (1² +³ dr) d=
=
dz
0
r² cos² sin 0r de dr dz
dr) dz
L² (²² + 1)² dz
6
1 ² (2²¹ +22² + 1) dz =
206
45
(4)
(4)
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