Calculus Please Help4

Consider the area between the graphs x+2y=9 and x+6=y^2. This area can be computed
in two different ways using integrals

First of all it can be computed as a sum of two integrals as

integral f(x)dx from a,b + integral g(x)dx from b,c

I got a=-6
b=3
c=19

but what does
f(x)=?
g(x)?

I'm thinking it is sqrt(x+6) and (9-x)/2 but it tell me I am getting it wrong.

Ty for your help!

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  1. You appear to have found their intersection correct as
    (3,3) and (19,-5)

    USING VERTICAL SLICES:
    y^2 = x+6
    so y = ± √(x+6) = ± (x+6)^(1/2)

    so for the part from -6 to 3
    height = (x+6)^ - ( -(x+6)^(1/2)
    = 2(x+6)^(1/2)
    integral of that is (4/3)(x+6)^(3/2)
    area of part 1
    = (4/3)(9)^(3/2) - (0)
    = 36
    2nd part of the region:
    heigh = (1/2)(9-x) - (-(x+6)^(1/2) )
    = (1/2)( (9-x) + (x+6)^(1/2) )
    integral of that is
    (1/2) ( 9x - x^2/2 + (2/3)(x+6)^(3/2) )
    evaluated from 3 to 19
    = (1/2) [ 171 - 361/2 + (2/3)(125) - (27 - 9/2 + (2/3)(27) ]
    = 50/3

    better check my arithmetic, I was too lazy to write it out first.
    So the area of the enclosed region using vertical slices = 36 + 50/3 = 158/3

    USING HORIZONTAL SLICES:
    width of a slice = 9-2y - (y^2 - 6)
    = 15 - 2y - y^2
    Area = ∫(15-2y-y^2) dy from -5 to 3
    = [15y - y^2 - y^3/3] from -5 yo 3
    = 45 - 9 - 27/3 - (-75 - 25 + 125/3)
    = 256/3

    ARGGG!! , error somewhere , sorry, perhaps you can spot it .

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