Expert: Paul Klarreich - 12/28/2007

Question
Let R be the region in the first quadrant under the graph of y = 1/square root of x for 4 <or equal to x <or equal to 9.
a. Find the area of R.
b. If the line x = k divides the region R into two regions of equal area, what it is the value of k?
c. Find the volume of the solid whose base is the region R and whose cross sections cut by planes
perpendicular to the x-axis are squares.

Answer
Questioner:   Stephano
Category:  Calculus
Private:  No
 
Subject:  AP Calculus
Question:  Let R be the region in the first quadrant under the graph of y = 1/sqrt(x) for 4 <= x <= 9.
a. Find the area of R.
b. If the line x = k divides the region R into two regions of equal area, what  is the value of k?
c. Find the volume of the solid whose base is the region R and whose cross sections cut by planes
perpendicular to the x-axis are squares.
 
 
a. That should be:
{9
|  x^(-1/2) dx =
}4
x^1/2
----- =
1/2

2 sqrt(x),  x = 4 to 9, =

2sqrt(9) - 2 sqrt(4) = 2

.......................................

b. That would be an equation like:


{k
|  x^(-1/2) dx = 1
}4

2sqrt(k) - 2 sqrt(4) = 1

2sqrt(k) - 4 = 1

2sqrt(k) = 5

sqrt(k) = 5/2

k = 25/4 = 6.25

c. Your 'typical' volume element has area  [f(x)]^2 = 1/x and thickness dx. So you want this integral:

{9
|   (1/x) dx = ln x, from 4 to 9 =
}4

ln 9 - ln 4