score update (after corrections)

Apologies & correction

I went home yesterday afternoon after classes, seminars and meetings and got pukey-sick, which is hopefully the only cause of my stupidity on Friday going over the quiz, for which I apologize and which I'm going to correct here.  If you still are confused on this issue, I'll be in my office tomorrow 11:30-12:30pm and Tuesday 12:30-1:30pm to discuss with you

The second problem on the quiz was to sketch out the regions of positive and negative divergence, which are as follows (blue is positive divergence and red is negative divergence)--I have imposed a coordinate system on the graph to give a language to describe parts of the graph:

Here's the correct reasoning. Calling the vector field F=Pi +Qj , the divergence is ∂P/∂x+∂Q/∂y; we can see that the Q is positive above the line y=1 and also below the line y=-1, and increasing as |y|>1 is more positive; you can see this in the way the vectors point upward-ish and are increasingly upward on each vertical line for x big enough or small enough.   On the other hand Q is negative between y=1 and y=-1, as you can see from the fact that the vectors point slightly downward.  Likewise, for P is positive for x bigger than 1 and also for x less than -1, and is more positive the more |x| is greater than 1 and is negative while x is  between x=-1 and x=1; you can see this from the way that the vectors point somewhat to the left when |x|<1 and to the right when |x|>1, and point increasingly to the right along horizontal lines as |x| is large enough.  

Now, since div(F)=∂P/∂x+∂Q/∂y, div(F) will be positive if the sum of ∂P/∂x and ∂Q/∂y is positive, that is if the rate of increase of P in the positive x-direction is greater than the rate of decrease of Q in the positive y-direction OR vice versa.  In the first quadrant, we can see that both ∂P/∂x and ∂Q/∂y are positive, in that they both P and Q increase from negative to positive values going from left to right (P) and from down to up (Q): this means div(F) must be positive.  The reverse is true in the third quadrant, so  div(F) must be negative.  In the half of the second quadrant above the line y=-x, Q increases a lot with increasing y while P decreases slightly with increasing x so ∂P/∂x is less negative than ∂Q/∂y is positive so div(F) must be positive; the opposite is true in the second quadrant below the line.  In the same manner, in the fourth quadrant above the line P is increasing quite a bit moving from left to right, while Q decreases not so much--it follows that ∂P/∂x is more positive than ∂Q/∂y is negative so div(F) must be positive, and again in the same manner the opposite is true in the fourth quadrant below the line.

Office Hours Monday and Tuesday

I announced this in class, but here it is anyway:
M 11:30AM-12:30PM
T 12:30-1:30PM

Finals Cram and Finals Workshop

From Brian England: I have decided to hold another problem session outside of the normal time frames next week.  I get out of my own final exam <<on Tuesday>> at 2pm and will head to the engineering tutoring center and be there by 2:30.  They have agreed to keep the place open until 5pm.  As always, your students are welcome to come.

Also: There will be a MAT267 Finals workshop Tuesday May 3 at from Noon to 1:00pm in BA201.  "Finals Workshops requires the student to bring class notes, practice material and questions to the workshop. Students should be prepared to collaborate with peers and tutors. This learning environment should foster recall and understanding of information for an upcoming exam."

  See <https://tutoring.asu.edu/calendar/>, Sign up: http://bit.ly/MAT267Final

13.7no7

I don't know why its coming out wrong!

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A few reasons.  First the vector element of area dA here is what our book calls dS, and not what the book calls dA--that is dA=(r_u x r_v )dudv, or since your parameters are called  r, θ  
dA=(r_r x r_θ )drdθ.  Second,  r, θ here are just parameters, not polar coordinates: you only have drdθ for your area element in the parameter plane, not rdrdθ (in fact I guess that the whole point of this problem is to fool you into making the mistake you did make, so that you can learn not to make it).  Finally, in your integral ∫∫_S F.dA it looks like you dropped the 9 from your F and also did the dθ integral incorrectly: the answer is not 2π because the F . (r_r x r_θ ) actually not independent of θ.

13.7no2

Can you help me with Let S be the part of the plane 3x+2y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=1i+1j+4k across the surface S.

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OK, here's an answer in general terms, so I don't do your homework problem for you.  You can write the plane as the graph z=f(x,y) of the function f(x,y)=4-3x-2y.  You can then use the usual parameterization of the graph of a function: r(u,v)=<u,v,4-3u-4v>, and then compute that
  r_u x r_v = ai+bj+ck, 
for whatever numbers a,b,c you get from computing the cross product.  Since the orientation is upward, you have to be a little careful: c has to be positive, so if   r_u x r_v  has c negative you have to use the -(r_u x r_v) instead to get the upward orientation.  The shadow of S in the xy-plane is the triangle enclosed by the x-axis, the y-axis, and the line 4-3x-2y=0, and the same triangle is the domain D of the parameterization. Then ∫∫_S F.dS=∫∫_D ∫∫_S F.(r_u x r_v) dA

13.6 no 3

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OK, the problem here is a practical one.  The "flow outward" this problem wants is in the units kg/sec. Since the density is kg/m^3,  and since the vector field v is in m/sec, and since dS has units m^2, the product density*v.dS has units of kg/sec.  Thus the rate of flow outward will be

5∫v.dS=5x(8/3)pi

 or (40/3)Pi.