Since S is part of the paraboloid z = g(x, y) = 3 - x² - y2, it is the graph of a function, and so we know that ag = 166 (-000 - 200 + R) DA. ду 16 Thus, we have the following. Step 3 Finally, F. ds = JS [²6² -xy(-2x) - yz(-2y) + zx da zx] Jo Jo = √₁*√² [2x²y + 2y²(3 − x² - y²) + x(3 − x² − y²)] dA = [² [²* (2x²y + 6₁,³² – 2x²,² – 2y² + 3x - − − x³ Step 2 Now, we have the following. 6² (2x²y + F. ds = 8x · + 6y² – 2x²y² – 2y4 + 3x − x³ – xy²) dy = x² + 3² √ ²₁ ( 1² × ² + + ² x - x ³ + ²/3) ∞ x = [ dx 6y² - 2y²x²2y² + 3x − x³ - xy²) dy dx +3 ∞ 100 8 + 8
Since S is part of the paraboloid z = g(x, y) = 3 - x² - y2, it is the graph of a function, and so we know that ag = 166 (-000 - 200 + R) DA. ду 16 Thus, we have the following. Step 3 Finally, F. ds = JS [²6² -xy(-2x) - yz(-2y) + zx da zx] Jo Jo = √₁*√² [2x²y + 2y²(3 − x² - y²) + x(3 − x² − y²)] dA = [² [²* (2x²y + 6₁,³² – 2x²,² – 2y² + 3x - − − x³ Step 2 Now, we have the following. 6² (2x²y + F. ds = 8x · + 6y² – 2x²y² – 2y4 + 3x − x³ – xy²) dy = x² + 3² √ ²₁ ( 1² × ² + + ² x - x ³ + ²/3) ∞ x = [ dx 6y² - 2y²x²2y² + 3x − x³ - xy²) dy dx +3 ∞ 100 8 + 8
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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![Step 1
Since S is part of the paraboloid z = g(x, y) = 3 - x² - y2, it is the graph of a function, and so we know that
ag
1/₂ +-08-16 (-0²² - Q² + ²) αa.
F
=
R)
R
dA.
ду
Thus, we have the following.
1 1
JJ
16 F
[*¹ [*¹
10
Step 3
F. ds =
Finally,
Step 2
Now, we have the following.
−xy(−2x) − yz(−2y) + zx] da
1 1
=
√ √ ² [²x²y + 2y²(3 − x² - y²) + x(3 − x² − y²)] a
dA
1
= [² [*¹ (2x²y + 6₁² – 2x²y³² – 2y4 + 3x
√² (2x²y + 6y²– 2x²y² - 2y² + 3x − x³ – xy²) dy =
6²³ ( 3² × ² + 1/ Xx - x ³ + ³/²-) d²
dx =
-
[
+
co
6y²2y²x²2y¹ + 3x - x³
8
-X
+
8
5
xy²) dy dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fccaac796-5fe2-4e0d-83bb-32b376fdab34%2F967da914-dfb5-4c54-88eb-ac7b51bbc663%2Frx09jj_processed.png&w=3840&q=75)
Transcribed Image Text:Step 1
Since S is part of the paraboloid z = g(x, y) = 3 - x² - y2, it is the graph of a function, and so we know that
ag
1/₂ +-08-16 (-0²² - Q² + ²) αa.
F
=
R)
R
dA.
ду
Thus, we have the following.
1 1
JJ
16 F
[*¹ [*¹
10
Step 3
F. ds =
Finally,
Step 2
Now, we have the following.
−xy(−2x) − yz(−2y) + zx] da
1 1
=
√ √ ² [²x²y + 2y²(3 − x² - y²) + x(3 − x² − y²)] a
dA
1
= [² [*¹ (2x²y + 6₁² – 2x²y³² – 2y4 + 3x
√² (2x²y + 6y²– 2x²y² - 2y² + 3x − x³ – xy²) dy =
6²³ ( 3² × ² + 1/ Xx - x ³ + ³/²-) d²
dx =
-
[
+
co
6y²2y²x²2y¹ + 3x - x³
8
-X
+
8
5
xy²) dy dx
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