Calculus
posted by Michael .
find the region enclosed by the given curves:
4x+y^2=9, x=2y

I assume you want the area of the region
you will need their intersection
use substitution ....
4(2y) + y^2 = 9
y^2 + 8y  9 = 0
(y1)(y+9) = 0
y = 1 or y 9
if y = 1 , x = 2 > (2,1)
if y = 9, x = 18 > (18, 9)
use horizontal slices
x = (1/4)y^2 + 9/2
the value of x for the region
= (1//4)y^2 + 9/2  2y
area = ∫(1/4 y^2 + 9/2  2y ) dy from 9 to 1
= [(1/20)y^5 + (9/2)y  y^2 ] from 9 to 1
= (1/20) + 9/2  1  ( (1/20)(59049) + 81/2  18)
= ....
you do the button  pushing.
(you might also want to check my arithmetic, my weakness)
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