Question

Consider the area between the graphs x+6y=8 and x+8=y2. This area can be computed in two different ways using integrals

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

where a= , b=, c= and
f(x)=
g(x)=

I found a, but not b or c. I can't seem to figure out f(x) and g(x). I would assume f(x) is (8-x)/6 and g(x) is (x+8)^(1/2), but it's not right.. Can you help me?

Answer 1

Here is a sketch of the region
http://www.wolframalpha.com/input/?i=solve+x%2B6y%3D8+%2C+x%2B8%3Dy%5E2

I will do it the easiest way.

The intersection points are (-4,2) and (56,-8)

I would take horizontal slices
effective width of a horizontal slice
= 8-6y - (y^2 - 8
= 16 - 6y - y^2

Area = ∫(16 - 6y - y^2) dy from y=-8 to 2
= (16y - 3y^2 - (1/3)y^3) | from -8 to 2
= (32 - 12 - 8/3) - (-128 - 192 + 512/3)
= 500/3 or
166 2/3

check my arithmetic