Suppose we have the open surface S, the upper hemisphere (with no bottom): z = f (x, y) = /1 – x² – y². If the fluid velocity field is F (x, y, z) = (y, –x, z²), then evaluate the polar double integral that is equivalent to: Net Upward Flux = // (F î) dS = ||(-Mfz– Nfy+P) dA, S R where în is the unit upward normal to S, and the region R is the projection of S down onto the xy-plane.
Suppose we have the open surface S, the upper hemisphere (with no bottom): z = f (x, y) = /1 – x² – y². If the fluid velocity field is F (x, y, z) = (y, –x, z²), then evaluate the polar double integral that is equivalent to: Net Upward Flux = // (F î) dS = ||(-Mfz– Nfy+P) dA, S R where în is the unit upward normal to S, and the region R is the projection of S down onto the xy-plane.
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![Net Upward Flux.
Suppose we have the open surface S, the upper hemisphere (with no bottom):
z = f (x, y) = /1 – x² – y?.
If the fluid velocity field is F (x, y, z) = (y, –x, z2), then evaluate the polar double integral
that is equivalent to:
-
Net Upward Flux =
(F.î) dS =
(-M fr – Nfy + P) dA,
S
R
where în is the unit upward normal to S, and the region R is the projection of S down onto
the xy-plane.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfe80350-6fa1-469d-aae2-95b4dd3aa499%2F77f42684-50db-4d02-8280-8c6051487742%2Ftigj1le_processed.png&w=3840&q=75)
Transcribed Image Text:Net Upward Flux.
Suppose we have the open surface S, the upper hemisphere (with no bottom):
z = f (x, y) = /1 – x² – y?.
If the fluid velocity field is F (x, y, z) = (y, –x, z2), then evaluate the polar double integral
that is equivalent to:
-
Net Upward Flux =
(F.î) dS =
(-M fr – Nfy + P) dA,
S
R
where în is the unit upward normal to S, and the region R is the projection of S down onto
the xy-plane.
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