Evaluate the surface integral [[F F. ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xy i+ yz j + zx k S is the part of the paraboloid z = 5 - x² - y² that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation Step 1 Since S is part of the paraboloid z = g(x, y) = 5 - x² - y², it is the graph of a function, and so we know that ag ag 16² - 16 (-38 - Q3% +R) dA. Əy F. ds = Thus, we have the following. 16² · [₁6² [-xy(-2x) - Yz(-2y) + 2x] da dA F. ds = - [² (² [2x²y + 2y ² (5 - x² - y²³) + x(5 − x² − y²)] da · − = - 6'²6² (2x²y + [ - xy²) dy dx
Evaluate the surface integral [[F F. ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xy i+ yz j + zx k S is the part of the paraboloid z = 5 - x² - y² that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation Step 1 Since S is part of the paraboloid z = g(x, y) = 5 - x² - y², it is the graph of a function, and so we know that ag ag 16² - 16 (-38 - Q3% +R) dA. Əy F. ds = Thus, we have the following. 16² · [₁6² [-xy(-2x) - Yz(-2y) + 2x] da dA F. ds = - [² (² [2x²y + 2y ² (5 - x² - y²³) + x(5 − x² − y²)] da · − = - 6'²6² (2x²y + [ - xy²) dy dx
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
![Evaluate the surface integral
find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x, y, z) = xy i+ yz j + zx k
S is the part of the paraboloid z = 5 x² - y² that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has
upward orientation
'//'
F. ds for the given vector field F and the oriented surface S. In other words,
Step 1
Since S is part of the paraboloid z = g(x, y) = 5 – x² - y², it is the graph of a function, and so we know that
ag ag
1/₂ - 0 ² - // (-²² - 0 + ₁)²
ds =
-P-
Q-
əx Əy
+R) da.
Thus, we have the following.
1
1
To
L'₁ [-xY(-2x) - 1
-xy(-2x) − yz(-2y) + zx da
-
[[ F
F. ds =
"1
=
- S₁²₁² [²2x²2y + 2y ² (5 − x² − y²) + x(5 − x² − y²)] dª
'1
= S'^ S^ (²x²y +
-
- xy²) dy dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa80fd0a7-9ae8-4bc5-b926-d69bfc10c0d7%2Fa180c4db-4df8-4943-91f9-2b3787052558%2F5hh5rkbp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Evaluate the surface integral
find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x, y, z) = xy i+ yz j + zx k
S is the part of the paraboloid z = 5 x² - y² that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has
upward orientation
'//'
F. ds for the given vector field F and the oriented surface S. In other words,
Step 1
Since S is part of the paraboloid z = g(x, y) = 5 – x² - y², it is the graph of a function, and so we know that
ag ag
1/₂ - 0 ² - // (-²² - 0 + ₁)²
ds =
-P-
Q-
əx Əy
+R) da.
Thus, we have the following.
1
1
To
L'₁ [-xY(-2x) - 1
-xy(-2x) − yz(-2y) + zx da
-
[[ F
F. ds =
"1
=
- S₁²₁² [²2x²2y + 2y ² (5 − x² − y²) + x(5 − x² − y²)] dª
'1
= S'^ S^ (²x²y +
-
- xy²) dy dx
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