Attendance Question #1

Use the divergence theorem to find IntInt(F*dS) if r[i] is the i-th digit of your RUID (if 0 make 1) 
F=<r[3]*x-r[5]*y+z, r[6]*x+z, x+r[9]*z>
Let S be the surface of the cube [0,3]x[0,4]x[0,5] 
With the normal pointing inwards

Answer to Attendance Question #1

F=<2*x-1*y+z, 2*x+z, x+3*z>
div(f)= 2+0+3= 5
Int(int(int(5)))= 5 x area
3x4x5= 60 
5 x 60 = 300
Inward means multiply by -1
-300

Attendance Question #2

By converting to polar coordinates

Int(from x=0 to x=3) int(y=-sqrt(9-x^2) to y=0 of (x^2+y^2)^3*x*y dy dx

Answer to Attendance Question #2

0<=x<=3 -sqrt(9-x^2)<=y<=0
-y=sqrt(9-x^2)
Y^2=sqrt(9-x^2)
X^2+y^2=9
R^2= 9
R=3

Int(Int(r^6*r^2costhetasintheta)) r*dr*dtheta
Int from 0 to 2*pi(int from 0 to 3(r^9sinthetacostheta))dr*dtheta
R^10/10 sintheta costheta from 0 to 3
Int(59049/10*sinthetacostheta) dtheta
59049/20*sintheta^2 from 0 to 2*pi
59049/20*0 = 0

Attendance Question #3

Investigate whether the following limits exist
(I) using the experimental approach 
(Ii) fully mathematical approach

Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (1,1,1)

Answer to Attendance Question #3

Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (1,1,1)
=  (1+2*1+3*1-6) / (3*1+2*1+1-6) 
=0/0

(I) Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (1,1,.9999)
= (1+2*1+3*.9999-6) / (3*1+2*1+.9999-6)
=3
Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (1,1,1.0001)
= (1+2*1+3*1.0001-6) / (3*1+2*1+.0001-6)
=3

(Ii) Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (1,y,1)
(1+2*y+3*1-6) / (3*1+2*y+1-6)= 2*y-2/2*y-2 = 1/1

Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (x,1,1)
(x+2*1+3*1-6) / (3*x+2*1+1-6)= x-1/3*x-3= 1/3

Limit of (x+2*y+3*z-6) / (3*x+2*y+z-6) as (x,y,z) to (1,1,z)
(1+2*1+3*z-6) / (3*1+2*1+z-6)=3*z-3/z-1= 3