#ATTENDANCE QUIZ for Lecture 12 of Math251(Dr. Z.)
#EMAIL  RIGHT AFTER YOU WATCHED THE VIDEO
#BUT NO LATER THAN  Oct. 19, 2020, 8:00PM (Rutgers time)
#THIS .txt FILE (EDITED WITH YOUR ANSWERS)
#TO:
#DrZcalc3@gmail.com
#Subject: aq12
#with an ATTACHMENT CALLED:
#aq12FirstLast.txt
#(e.g. aq12DoronZeilberger.txt)




#LIST ALL THE ATTENDANCE QUESTIONS FOLLOWED BY THEIR ANSWERS


Question 1. 
Let a[I] be the I-th digit of your RUID (if it is 0 make it 1)
Let c=min (a[I],a[5]); d=max(a[1],a[5])
Let f(x)= a[2]*x^2+a[3]
Find the area bounded by x=c and x=d above the x-axis

Answer to Question 1. 
With my RUID, f(x)  = (9*x^2+6)
Maximum y value o ff(x) on 1 to 5 is y=81

set top double integral:
Integral of y=0 to y=81, Integral of x=1 to x=5 (9*x^2+6) dx dy

Evaluate inner integral first= 366

Evaluate outer integral: 29,646

Answer: Area bounded by x=1 to x=5 above the x-axis and f(x) is 29,646 units^2. 
 

Question 2. 
Find the double integral of f(x,y)= a[5]*x+a[7]*y+a[1]
R= { (x,y): 0<=x<=a[9], 1<=y<=a[6] }


IMPORTANT SHORTCUT: IF THE INTEGRAND IS 1 f(x,y)=c THEN IntInt( f(x,y), OVER R)= c*AREA(R)

Answer to Question 2. 

With my RUID: f(x,y) = x + 7*y + 1

Set up integral: 

Integral of y=0 to y=1, Integral of x=0 to x=2 (x + 7*y + 1) dx * dy

Evaluate first integral = 2+14y+2

Evaluate outer integral=11

Answer: The value of the double integral is 11. 



Question 3. 
Find the integral of f(x,y)= sin(x+y) over the region R= ( 0<=x<=pi/2, 0<=y<=pi/2)

Answer to Question 3. 

Set up the integral:

Integral of y=0 to y= pi/2, Integral of x=0 to x= pi/2 ( sin(x+y)) dx* dy

Evaluate inner integral= -cos( pi/2 +y) + cos(y)

Evaluate outer integral= 2

Answer: The integral is equivalent to 2.