the final

just in case you get lost, the following is a screen shot from the syllabus.

A practice final exam for tomorrow

Practice Final

Here's a practice final. I suggest you take an hour to work the problems before you go to class tomorow.

Current scores and estimated grades

The estimated score is based on extrapolating your final exam score as the average of your exam scores to date. The estimated flat and curved grades are determined from the estimated score using according to the grade templates described in the course syllabus.

Link to the official final exam study materials

Practice Final (Do this part first)

Practice Final Solutions 

Review

13.4#5

Hello Professor,
You asked me to email you about this question so you could explain it on the blog.
Much Appreciated,


*********************************

Ah. This is a different problem than the one I thought we were talking about.   Notice that the problem says that the path is "nonclosed"?  This means that you don't have a closed loop, just the three line segments AB+BC+CD, but you don't have DA also, so Green's theorem doesn't directly apply.  You could solve this problem directly just by integrating F=<sin(x)+3y, 2x+y> and integrating
∫ F.dr_1 + ∫ F.dr_2 + ∫ F.dr_3 where r_1(t)=t*<2,2> for 0≤t≤1, r_2(t)=(1-t)*<2,2> + t*<2,4> for 0≤t≤1, r_3(t)=(1-t)*<2,4> + t*<0,6> for 0≤t≤1. Or you *could* still use Green's theorem if you're willing to be a little sneaky: since ∂F_2/∂x = 2 and ∂F_1/∂y = 3, Green's theorem implies that
∬_D (-1)dA - ∫ F.dr_4 = ∫ F.dr_1 + ∫ F.dr_2 + ∫ F.dr_3 ,  where D is the enclosed region and where r_4(t)=(1-t)*<0,6> . This is a little bit easier approach since ∬_D (-1)dA = -Area = -8, and
r'_4(t)= -<0,6> so ∫ F.dr_4 = ∫_0^1<18(1-t), 6(1-t)>.<0, -6> dt = -36 ∫_0^1(1-t) dt.

Lecture Notes 4/19/19 through 4/24/19 (to be completed)

Lecture Notes 4/19/19

Lecture Notes 4/22/19

Lecture Notes 4/24/19

Lecture Notes 4/3/19 through 4/17/19

Lecture Notes 4/3/19

Lecture Notes 4/5/19

Lecture Notes 4/8/19

Lecture Notes 4/10/19

Lecture Notes 4/17/19

PracticeTest3 #2

Hello professor,
I was working on the Exam 3 practice ( https://math.asu.edu/sites/default/files/mat_267-post_exam3_solutions_0.pdf ) and I was confused on question #2 of the multiple choice. For the limits of integration of phi I got found them to be from 0 to pi. However, the correct answer says its from 0 to pi/2. Why is that?

Thank you

********************

In Cartesian (xyz) coordinates, your integration dz integration runs from z=0 to z=√(4-x^2-y^2); i.e. from the xy-plane UP to the top half of the sphere x^2+y^2+z^2=4, specifically there's nothing below the xy-plane. But the xy-plane is at 90 degrees or π/2 radians from the positive z-axis, hence π/2 has to be the upper limit.  If you went all the way to π your domain would have to include the negative z axis. 

Lecture Notes 3-27-19 through 4-1-19 (to be updated)

Lecture Notes 3/27/19

Lecture Notes 3/29/19

Lecture Notes 4/1/19