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

13.7 no 7.

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 no3

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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.

Attendance to date

(please click on the image to enlarge)

Exam 3 Update Glitch

Just at the end of my office hour (and a half) today I crashed my spreadsheet.  I was saving often, so this could only possibly affect the people who were in my office after 12:30PM.  I *think* I recovered all of the correct scores, but if you were in my office after 12:30PM today, and correcting your exam or quiz 9 scores, check the following list.  If there is a problem, bring bring your additional scores (the new points I gave you on your paper) to my office hour tomorrow or class on Friday and I'll fix the problem

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