Calculus of a Single Variable, Chapter 7, 7.1, Section 7.1, Problem 70

Given ,
y = 9-|x| , y = 0
first let us find the total area of the bounded by the curves.
so we shall proceed as follows
as given ,
y = 9-|x| , y = 0
=> 9-|x|=0
=> |x| -9 =0
=> |x|=9
so x=+-9

the the area of the region is = int _-9 ^9 (9-|x| -0) dx
=>int _-9 ^0 (9+x -0) dx+int _0 ^9 (9-x -0) dx
=>[9x+x^2 /2]_-9 ^0 + [9x-x^2/2]_0 ^9
=>[0]-[-81+81/2] +[81-81/2]-[0]
=>81/2 +81/2 =81
So now we have to find the horizonal line that splits the region into two regions with area 81/2
as when the line y=b intersects the curve y=9-|x| then the area bounded is 81,so
let us solve this as follows
first we shall find the intersecting points
as ,
9-|x|=b
|x|= 9-b
x=+-(9-b)
so the area bound by these curves y=b and y=9-|x| is as follows
A= int _-(9-b) ^(9-b) (9-|x|-b)dx = 81/2
=>int _-(9-b) ^0 (9+x-b)dx +int _0 ^(9-b) (9-x-b)dx =81/2
=>[9x+x^2/2-bx]_-(9-b) ^0 +[9x-x^2/2-bx]_0 ^(9-b) = 81/2
=>[0]-[9(-(9-b))+(-(9-b))^2 /2-b(-(9-b))]+
[9((9-b))-((9-b))^2/2-b((9-b))]-[0]=81/2
=> [9((9-b))-((9-b))^2/2-b((9-b))]
-[9(-(9-b))+(-(9-b))^2/2 -b(-(9-b))]=81/2
let t= 9-b
so
=> [9(t)-(t)^2/2 -b(t)] -[9(-t)+(-t)^2/2 -b(-t)]=81/2
=>[9t-t^2/2 -bt]+[9t-t^2/2 -bt]=81/2
=>18t-t^2-2bt =81/2
but we know half the Area of the region between y=9-|x|,y=0 curves =81/2
so now ,
18t-t^2-2bt =81/2
18t-2bt-t^2 = 81/2
=>t(18-2b)-t^2=81/2
=> t^2=t(18-2b)-81/2
=>t^2 -t(18-2b)+81/2=0
this is like quadratic equation
ax^2+bx+c=0
so t = (-b+-sqrt(b^2-4ac))/2a
=(18-2b+-sqrt((-(18-2b))^2-4*(81/2)))/2
but
t=9-b
so,
9-b=(18-2b+-sqrt((-(18-2b))^2-4*(81/2)))/2
=> 18-2b=(18-2b+-sqrt((-(18-2b))^2-4*(81/2)))
=>+-sqrt((-(18-2b))^2-4*(81/2))=0
=>sqrt((-(18-2b))^2-4*(81/2))=0
=>(-(18-2b))^2-4*(81/2)=0

=>(-(18-2b))^2=4*(81/2)
=>(-(18-2b))^2=2*(81)
=>(-(18-2b))= +- sqrt(2) *9
=> -18+2b=+-9sqrt(2)
=>2b=+-9sqrt(2)+18
=>b=(18+-9sqrt(2))/2
so b= (18+-9sqrt(2))/2