Calculus II 
Quiz #5 
March 13, 2002
Name____________________    
R.  Hammack 
Score ______

(1) Find the area of the region contained between the graphs of   [Graphics:Images/quiz5sol_gr_1.gif] and [Graphics:Images/quiz5sol_gr_2.gif]

The region is sketched below. Notice that the curves intersect at (0,0) and (2,4), and  [Graphics:Images/quiz5sol_gr_3.gif]is the upper function on [0,2].

[Graphics:Images/quiz5sol_gr_4.gif]


The area is thus [Graphics:Images/quiz5sol_gr_5.gif](4-8/3)-(0-0/3) = 4/3 square units.

(2)  Consider region under the curve [Graphics:Images/quiz5sol_gr_6.gif] between x = 1 and x = 2.

(a) The region is revolved around the x-axis. Find the volume of the resulting solid.
The cross section at x is a circle of radius 1/x, so its area is A(x) = π[Graphics:Images/quiz5sol_gr_7.gif]π[Graphics:Images/quiz5sol_gr_8.gif]
Thus the volume of the solid is [Graphics:Images/quiz5sol_gr_9.gif]
= [Graphics:Images/quiz5sol_gr_10.gif] =   π/2 cubic units.


(b) The region is revolved around the y-axis. Find the volume of the resulting solid.
Using the method of shells, we get V=[Graphics:Images/quiz5sol_gr_11.gif]=[Graphics:Images/quiz5sol_gr_12.gif]=[Graphics:Images/quiz5sol_gr_13.gif]=[Graphics:Images/quiz5sol_gr_14.gif]
=4π-2π = 2π cubic units.