<p>I heard a rumor that it might not be. Can someone verify?</p>
<p>I think it’s cumulative, since the practice final is.</p>
<p>the 08 final actually wasn’t</p>
<p>in lecture today, he told us the types of problems that will be on the final and it’s pretty much just chapter 16.</p>
<p>Final Exam Questions:
- Line Integrals in plane (3 part problem; he said the 3rd part would be almost impossible to solve)
- fundamental theorem of cal for line integrals (2 or 3D)
- Simple essence of proof of green’s theorem
- calculation/application of Green’s theorem
- surface in R^3, normal vector, area r(u,v) or z=f(x,y)
- surface integral of a vector field (know definition and calculate correctly)
- divergence theorem
- stokes theorem</p>
<p>Oh snap! I stand corrected.</p>
<p>thanks Lupirius</p>
<p>did you guys get “1” for the limit of A/pi*r^2?</p>
<p>That’s what I got</p>
<p>i got that too - good 
what other answers did you guys get for the questions?
(i assume we are allowed to discuss this since we could keep the final)
and what did you think of the final in general?</p>
<p>you know for #1C, where it went from (0,0) to (1/2,1/2) to down the path of the ellipse back to the origin? that was a clockwise curve. could we just evaluate the line integral ydx directly? i just added those 3 partial line integrals up. wasn’t sure if we had to do something with the sign though, i coulda swore i heard my GSI saying something about having to flip the sign if it’s clockwise? i dunno.</p>
<p>Well, pretty much every problem could be solved 2 ways, since we had all those theorems. There’s the straightforward way (which made me feel dumb) and the smart way (which made me feel nervous about it being indirect). I generally did it one way and checked it with the other since hey, we had 3 hours.</p>
<p>The straightforward way is to add the three up, which isn’t too bad since the flat part is 0.
If you look at Green’s theorem it’s just the area of that section.
integral (-y dx) over counterclockwise, or integral (y dx) over clockwise.</p>
<p>I did the straightforward way - I think using Green’s Theorem would take more work. The radius goes from 0 to a function of theta, so that would be a little tedious.</p>
<p>The straightforward ydx integral involved copying down the same thing you did for 1a and 1b, replacing a couple numbers, and adding it to the simple integral of ydx for y=f(x)</p>
<p>2 questions:</p>
<p>(1) anyone know when grades are going up onto bearfacts? my gsi didn’t say anything, did yours? or did the prof say anything?</p>
<p>(2) is there any way to see our finals? do we go to neu’s office or something?</p>
<p>i did the adding up method too but I have no idea if it is right. the 3rd part is zero, but what did you guys get for the first and second parts? what integrals needed to be set up?</p>
<p>as for the final i have no clue…email the GSI i guess. i wish they’d post an answer key.</p>
<p>any answers for any other questions?</p>
<p>1a and 1b were the same thing, except I believe 1a was negative. Given x=cos t and y=1/rt3 sin t, dx=-sint dt
so fnInt (1/rt3 sint (-sint dt)), which integrates to 1/rt3 [t/2-sin 2t/4] over [a,b].</p>
<p>1c: Well, you do still need to find theta for the straightforward way. I’m just grateful because I made a mistake finding theta (i had put pi/4 at first), and after trying to reconcile both methods, realized it was wrong. The answer should have been pi/(4 rt 3) , I think.
The curved section should go from pi/3 to 0 (if 1b is from 2 pi to 0); the radial line is x=y so fnInt(x dx) over [0,.5] by substitution, and the flat line is 0. By integrating the curved section, you get a remainder of -1/8 which cancels with the 1/8 from the radial line.</p>
<p>My GSI said to e-mail him Monday for the final score, so you could try likewise. I don’t know if you can see your final. I hear it’s policy not to do regrades, anyhow. Only counting errors can be changed.</p>
<p>grades are up</p>
<p>is there anyway we can find out what we got on the final, or does bearfacts just post final grades in the course?</p>