`y = 9-x^2 , y = 0` Find b such that the line y = b divides the region bounded by the graphs of the equations into two regions of equal area.

Expert Answers
kseddy123 eNotes educator| Certified Educator

Given ,

`y = 9-x^2 , 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^2 , y = 0`

=>` 9-x^2=0`

=> `x^2 -9 =0`

=>` (x-3)(x+3)=0`

so `x=+-3`

 

the the area of the region is = `int _-3 ^3 (9-x^2 -0) dx`

=`[9x-x^3/3] _-3 ^3`

=` [27-9]-[-27+9]`

=`18-(-18) = 36`

So now we have  to find the horizonal line that splits the region into two regions with area 18

as when the line y=b intersects the curve `y=9-x^2` then the ared bounded is 18,so

let us solve this as follows

first we shall find the intersecting points

as ,

`9-x^2=b`

`x^2= 9-b`

`x=+-sqrt(9-b)`

so the area bound by these curves `y=b` and `y=9-x^2 ` is as follows

A= `int _-sqrt(9-b) ^sqrt(9-b) (9-x^2-b)dx = 18`

=> `int _-sqrt(9-b) ^sqrt(9-b)(9-x^2-b)dx=18`

=> `[-bx +9x-x^3/3]_-sqrt(9-b) ^sqrt(9-b)`

=>`[x(9-b)-x^3/3]_-sqrt(9-b) ^sqrt(9-b)`

=>`[((sqrt(9-b))*(9-b))-[(sqrt(9-b))^(3)]/3 ]-[((-sqrt(9-b))*(9-b))-[(-sqrt(9-b))^(3)]/3]`

=>`[(9-b)^(3/2) - ((9-b)^(3/2))/3]-[-(9-b)^(3/2)-(-((9-b)^(3/2))/3)]`

=>`[(2/3)[9-b]^(3/2)]-[-(9-b)^(3/2)+((9-b)^(3/2))/3]`

=>`(2/3)[9-b]^(3/2) -[-(2/3)[9-b]^(3/2)]`

=`(4/3)[9-b]^(3/2)`

but we know half the Area of the region between `y=9-x^2,y=0` curves =`18`

so now ,

`(4/3)[9-b]^(3/2)=18`

 

let `t= 9-b`

=> `t^(3/2)= 18*3/4`

=> `t=(27/2)^(2/3)`

=>` 9-b= 9/(root3 (4))`

 

=> `b= 9-9/(root3 (4))`

 

=`9(1-1/(root3 (4))) ` = `3.330`

 

so `b= 3.330`