The surface x = = √√2r cos 0, y = r sin 0, z = −r sin 0, 0 ≤ r≤ 1, 0 ≤ 0 ≤ 2π is an inclined disk of radius √2 and its boundary C may be parameterized by substituting r counterclockwise when viewed from above). Using Stokes' theorem, calculate M = F = (P, Q, R), P = 10xz-2y²+10x +4z, So C == 1 (and then increasing is F. dr for the vector field Q = - 4xy − 8z² + x + 3y + z, R = 5x2 The value of sin (1 + 5M¹/³) is 16yz + 2x - 4y. -0.094 -0.358 -0.019 -0.876 0.089 -0.557 0.285 0.104 .

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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The surface
x = = √√2r cos 0, y = r sin 0, z = −r sin 0, 0 ≤ r≤ 1, 0 ≤ 0 ≤ 2π
is an inclined disk of radius √2 and its boundary C may be parameterized by substituting r
counterclockwise when viewed from above). Using Stokes' theorem, calculate M =
F = (P, Q, R),
P = 10xz-2y²+10x +4z,
So
C
== 1 (and then increasing is
F. dr for the vector field
Q
=
- 4xy − 8z² + x + 3y + z,
R = 5x2
The value of sin (1 + 5M¹/³) is
16yz + 2x - 4y.
-0.094
-0.358
-0.019
-0.876
0.089
-0.557
0.285
0.104
.
Transcribed Image Text:The surface x = = √√2r cos 0, y = r sin 0, z = −r sin 0, 0 ≤ r≤ 1, 0 ≤ 0 ≤ 2π is an inclined disk of radius √2 and its boundary C may be parameterized by substituting r counterclockwise when viewed from above). Using Stokes' theorem, calculate M = F = (P, Q, R), P = 10xz-2y²+10x +4z, So C == 1 (and then increasing is F. dr for the vector field Q = - 4xy − 8z² + x + 3y + z, R = 5x2 The value of sin (1 + 5M¹/³) is 16yz + 2x - 4y. -0.094 -0.358 -0.019 -0.876 0.089 -0.557 0.285 0.104 .
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