Flux of the amount of buffalo entering a square kilometer per minute












1












$begingroup$


enter image description here



So Eq.(13) is just $int F.n dS = int int div(F) dA$



So I did $div(F) = nabla . F = y+1$, then I did $int_2^3 int_2^3 y + 1 dx dy = int_2^3 y + 1 dy = .5(3^2) + 3 - (.5(2^2) + 2) = 3.5$ This multiplied by density is 1750.



This is not the answer. The answer is 29.2










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$endgroup$












  • $begingroup$
    What book is this?
    $endgroup$
    – rogerl
    Apr 21 '16 at 0:43










  • $begingroup$
    Rogawski Calculus 3rd edition
    $endgroup$
    – user116528
    Apr 21 '16 at 0:45
















1












$begingroup$


enter image description here



So Eq.(13) is just $int F.n dS = int int div(F) dA$



So I did $div(F) = nabla . F = y+1$, then I did $int_2^3 int_2^3 y + 1 dx dy = int_2^3 y + 1 dy = .5(3^2) + 3 - (.5(2^2) + 2) = 3.5$ This multiplied by density is 1750.



This is not the answer. The answer is 29.2










share|cite|improve this question









$endgroup$












  • $begingroup$
    What book is this?
    $endgroup$
    – rogerl
    Apr 21 '16 at 0:43










  • $begingroup$
    Rogawski Calculus 3rd edition
    $endgroup$
    – user116528
    Apr 21 '16 at 0:45














1












1








1





$begingroup$


enter image description here



So Eq.(13) is just $int F.n dS = int int div(F) dA$



So I did $div(F) = nabla . F = y+1$, then I did $int_2^3 int_2^3 y + 1 dx dy = int_2^3 y + 1 dy = .5(3^2) + 3 - (.5(2^2) + 2) = 3.5$ This multiplied by density is 1750.



This is not the answer. The answer is 29.2










share|cite|improve this question









$endgroup$




enter image description here



So Eq.(13) is just $int F.n dS = int int div(F) dA$



So I did $div(F) = nabla . F = y+1$, then I did $int_2^3 int_2^3 y + 1 dx dy = int_2^3 y + 1 dy = .5(3^2) + 3 - (.5(2^2) + 2) = 3.5$ This multiplied by density is 1750.



This is not the answer. The answer is 29.2







multivariable-calculus vector-fields fluid-dynamics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 21 '16 at 0:24







user116528



















  • $begingroup$
    What book is this?
    $endgroup$
    – rogerl
    Apr 21 '16 at 0:43










  • $begingroup$
    Rogawski Calculus 3rd edition
    $endgroup$
    – user116528
    Apr 21 '16 at 0:45


















  • $begingroup$
    What book is this?
    $endgroup$
    – rogerl
    Apr 21 '16 at 0:43










  • $begingroup$
    Rogawski Calculus 3rd edition
    $endgroup$
    – user116528
    Apr 21 '16 at 0:45
















$begingroup$
What book is this?
$endgroup$
– rogerl
Apr 21 '16 at 0:43




$begingroup$
What book is this?
$endgroup$
– rogerl
Apr 21 '16 at 0:43












$begingroup$
Rogawski Calculus 3rd edition
$endgroup$
– user116528
Apr 21 '16 at 0:45




$begingroup$
Rogawski Calculus 3rd edition
$endgroup$
– user116528
Apr 21 '16 at 0:45










1 Answer
1






active

oldest

votes


















1












$begingroup$

The value 1750 corresponds with the number of buffalo entering/leaving the area per hour. Divide by 60 to get 29.2.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer is correct. It is asking for buffalo per minute and the flux is given per hour
    $endgroup$
    – whpowell96
    Dec 12 '18 at 19:28











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The value 1750 corresponds with the number of buffalo entering/leaving the area per hour. Divide by 60 to get 29.2.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer is correct. It is asking for buffalo per minute and the flux is given per hour
    $endgroup$
    – whpowell96
    Dec 12 '18 at 19:28
















1












$begingroup$

The value 1750 corresponds with the number of buffalo entering/leaving the area per hour. Divide by 60 to get 29.2.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer is correct. It is asking for buffalo per minute and the flux is given per hour
    $endgroup$
    – whpowell96
    Dec 12 '18 at 19:28














1












1








1





$begingroup$

The value 1750 corresponds with the number of buffalo entering/leaving the area per hour. Divide by 60 to get 29.2.






share|cite|improve this answer











$endgroup$



The value 1750 corresponds with the number of buffalo entering/leaving the area per hour. Divide by 60 to get 29.2.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 19:18









Nij

2,03011223




2,03011223










answered Dec 12 '18 at 17:51









David StevensDavid Stevens

111




111












  • $begingroup$
    This answer is correct. It is asking for buffalo per minute and the flux is given per hour
    $endgroup$
    – whpowell96
    Dec 12 '18 at 19:28


















  • $begingroup$
    This answer is correct. It is asking for buffalo per minute and the flux is given per hour
    $endgroup$
    – whpowell96
    Dec 12 '18 at 19:28
















$begingroup$
This answer is correct. It is asking for buffalo per minute and the flux is given per hour
$endgroup$
– whpowell96
Dec 12 '18 at 19:28




$begingroup$
This answer is correct. It is asking for buffalo per minute and the flux is given per hour
$endgroup$
– whpowell96
Dec 12 '18 at 19:28


















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