Question

find the area of the region:
y=4
y=2sqrt(x)
y=3-x

Answer 1

first two hit at (4,4)

curve 2 and 3 hit at
3 - x = 2 sqrt x
x + 2 sqrt x = 3
let z = sqrt x
z^2 + 2 z - 3 = 0
(z+3)(z-1) = 0
sqrt x = -3 or sqrt x = 1
so x = 1, y = 2

line 3 and line 1 hit at
(-1, 4)
draw all that on your sketch
now from x = -1 to x = 1
integrate [ 4 - 3+x) or in other words
(1+x) dx
and from x = 1 to x = 4
integrate [ 4 - 2 x^.5) dx
and add the two results

Answer 2

or, you can integrate over dy, and avoid the split:

In this case, the two lower bounding curves are
x = y^2/4
x = 3-y

∫[2,4] y^2/4 -(3-y) dy = 14/3

Checking with Damon's method gives

∫[-1,1] 4-(3-x) dx + ∫[1,4] 4-2√x dx
= 2 + 8/3
= 14/3

So, each method has its complications.