Consider the surface, S, given by the portion of the parabaloid z = 14-4x² – 2y² above the xy-plane (where z > 0). A flow field is given by F = <7y, 6x, 5xy>. Which of the following integrals gives the flux across the surface S? A) [ƒ (7y+6x)√1+x²+y²d$_(B) [√ √1+64x²+16y²dS (C) S [dS (H) √√ √ ₁ +87² + 4y=²0 7y + 6x + 5xy dS S -6 √1+49x²+36 85xy _D) [ƒ 40 √/1 +64x²+16y² dS (E) [ƒ (14−4x²−29²)√1 + 16x² +41³² dS (F) [√ √1+64x² +16² dS S S S 14 _G) [√ √₁+1687² +4² S dS

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter7: Integration
Section7.5: The Area Between Two Curves
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Problem #9: Consider the surface, S, given by the portion of the parabaloid = = 14-4x² -22 above the xy-plane (where z >
0). A flow field is given by F = <7y, 6x, 5xy>. Which of the following integrals gives the flux across the surface
S?
-6
IdS (B) ƒƒ √1+64x² +16y²dS (C) [√ √/1+49x² +36²
dS
S
S
(A)
· ƒƒ (7y+6x)√1 + x² + y²dS (B) [f
85xy
(D) ƒƒ˜ 40 √/1 +64x²+ 16y² dS (E) ƒƒ (14−4x² −2y³²)√1+16x² + 4y² d$ (F) √ √ √₁ +64x² + 16y²²
dS
S
(G) SS √1+16x² + 4y² 5dS (H) SS √ ₁ + 8&x² + 4 ² ²
14
7y + 6x + 5xy
5dS
S
Transcribed Image Text:Problem #9: Consider the surface, S, given by the portion of the parabaloid = = 14-4x² -22 above the xy-plane (where z > 0). A flow field is given by F = <7y, 6x, 5xy>. Which of the following integrals gives the flux across the surface S? -6 IdS (B) ƒƒ √1+64x² +16y²dS (C) [√ √/1+49x² +36² dS S S (A) · ƒƒ (7y+6x)√1 + x² + y²dS (B) [f 85xy (D) ƒƒ˜ 40 √/1 +64x²+ 16y² dS (E) ƒƒ (14−4x² −2y³²)√1+16x² + 4y² d$ (F) √ √ √₁ +64x² + 16y²² dS S (G) SS √1+16x² + 4y² 5dS (H) SS √ ₁ + 8&x² + 4 ² ² 14 7y + 6x + 5xy 5dS S
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