[15] (4) GIVEN: a > 0, a constant, and the directed path C, (Add extra pages, as needed. from A = (a,0) to B= (0, a), directed along the line segment, AB; and, the vector field in the xy - plane, F = (x² - y², x² + y²). FIND: The path integral: F•ds (0, a) / 1 1 ΚΑ (a,0) / X 1 1 1 F = (x² − y², x² + y²)
[15] (4) GIVEN: a > 0, a constant, and the directed path C, (Add extra pages, as needed. from A = (a,0) to B= (0, a), directed along the line segment, AB; and, the vector field in the xy - plane, F = (x² - y², x² + y²). FIND: The path integral: F•ds (0, a) / 1 1 ΚΑ (a,0) / X 1 1 1 F = (x² − y², x² + y²)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
for the first image attach please do the calculations similar to the second image attach
please please answer every correctly also this is not a graded question I would really appreciate if you would answer
![[15] (4) GIVEN: a > 0, a constant, and the
directed path C,
from A = (a,0) to B= (0, a),
directed along the line segment, AB;
Add extra
pages, as
needed.
and, the vector field in the xy - plane,
(x² − y², x² + y²).
F
FIND: The path integral: F•ds
=
6 77
1
A
(a,0) /
1
17
F = (x² − y², x² + y²)
1
1
1
A 1
1
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe343d170-4423-4dcd-9c75-f5f0118e0ff9%2F3c6c6bd4-6053-421b-ae71-45b633229c68%2Fczmtutf_processed.png&w=3840&q=75)
Transcribed Image Text:[15] (4) GIVEN: a > 0, a constant, and the
directed path C,
from A = (a,0) to B= (0, a),
directed along the line segment, AB;
Add extra
pages, as
needed.
and, the vector field in the xy - plane,
(x² − y², x² + y²).
F
FIND: The path integral: F•ds
=
6 77
1
A
(a,0) /
1
17
F = (x² − y², x² + y²)
1
1
1
A 1
1
X
![[15] (4)
GIVEN: a > 0, a constant, and the
directed path c,
from A = (a,0) to B= (a, a) to C = (0,0),
directed along the line segments, AB, BC;
and, the vector field in the xy - plane,
F = (x + y, - x).
FIND: The path integral: F·ds
METHOD:
1
AB: C = (a,y), ye [o,a]
BC: -C₂ = (x,x), z€ [0,a]
di
F•
=
Lên đã đến đâện đã
-C₂
.
Lên đã = f(a+y, a)(0,1)
= √² (-a) dy = − a²
Hence,
(a,0)
F = (x+y₁-x)
·a
d
√ F• dî = [ª (2x,-z). (1,1) 4 = √x+
-C₂
= 1/α²
√ F• dª = -a² - £a² = −2a²
The negative
value of the answer,
(checks with the sketch.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe343d170-4423-4dcd-9c75-f5f0118e0ff9%2F3c6c6bd4-6053-421b-ae71-45b633229c68%2Fmftzi5_processed.png&w=3840&q=75)
Transcribed Image Text:[15] (4)
GIVEN: a > 0, a constant, and the
directed path c,
from A = (a,0) to B= (a, a) to C = (0,0),
directed along the line segments, AB, BC;
and, the vector field in the xy - plane,
F = (x + y, - x).
FIND: The path integral: F·ds
METHOD:
1
AB: C = (a,y), ye [o,a]
BC: -C₂ = (x,x), z€ [0,a]
di
F•
=
Lên đã đến đâện đã
-C₂
.
Lên đã = f(a+y, a)(0,1)
= √² (-a) dy = − a²
Hence,
(a,0)
F = (x+y₁-x)
·a
d
√ F• dî = [ª (2x,-z). (1,1) 4 = √x+
-C₂
= 1/α²
√ F• dª = -a² - £a² = −2a²
The negative
value of the answer,
(checks with the sketch.
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