Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xzey i − xzey j + z k S is the part of the plane x + y + z = 3 in the first octant and has upward orientation Incorrect: Your answer is incorrect. 9/2
1 answer:
I'm pretty sure your answer of 9/2 is correct...
Parameterize by
with and .
In case youre' not sure where this came from: recall the standard parameterization for a line segment connecting two points and ,
with . Then treating as a "point", we can parameterize a plane containing and another point by using
with .
Now, take the normal vector to to be
Then the flux of across is
(since the first two components of cancel in the dot product)
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First you have to change each one into a decimal. 10% = 0.10 1/9 = 0.11 So the answer has to be between 0.10 and 0.11. A is more than 0.11. B is more than 0.11. C is between 0.10 and 0.11. D is less than 0.10. The answer is C. 0.108.
Answer:
.25 inches per hour
Step-by-step explanation:
.5 ft
-----------
day
We need to get to inches
1 ft = 12 inches
.5 ft 12 inches
----------- * ------------
day 1 ft
We also need to convert days to minutes
1 day = 24 hours
.5 ft 12 inches 1 day
----------- * ------------ * -------------
day 1 ft 24 hours
6 inches
-------------
24 hours
.25 inches per hour
the answer is 32 because you add 9 plus 7 then you multiply tha th # by 2
About three times. Just divide 74 by 20 which is 3 but exactly it is 3.7. hoped it helped
Answer:
Step-by-step explanation: