Sum of two Integrals

First plot the curves:
http://www.wolframalpha.com/input/?i=plo…

Notice that leftmost point is vertex of parabola (−8, 0)
The curves intersect at (17, 5) and (56, −8)

Since we are using two integrals, there will not be just one f(x) and one g(x)

On interval −8 < x < 17, top function is top of parabola
and bottom function is bottom of parabola:
f(x) = √(x+8)
g(x) = −√(x+8)

On interval 17 < x < 56, top function is line
and bottom function is botto of parabola:
f(x) = (32−x)/3
g(x) = −√(x+8)

Area = ∫₋₈¹⁷ (√(x+8) − −√(x+8)) dx + ∫₁₇⁵⁶ ((32−x)/3 − −√(x+8)) dx
Area = ∫₋₈¹⁷ 2√(x+8) dx + ∫₁₇⁵⁶ (32/3 − x/3 + √(x+8)) dx
Area = 4/3 (x+8)^(3/2) |₋₈¹⁷ + (32/3 x − x²/6 + 2/3 (x+8)^(3/2)) |₁₇⁵⁶
Area = 4/3 (25^(3/2) − 0^(3/2)) + 32/3 (56−17) − (56²−17²)/6 + 2/3 (64^(3/2)−25^(3/2))
Area = 4/3 (125) + 32/3 (39) − 2847/6 + 2/3 (387)
Area = 2197/6
Area = 366 ⅙

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Alternate method is to integrate with respect to y
This will require 1 integral only

On interval −8 < y < 5
upper function is line -----> f(y) = 32 − 3y
lower function is parabola ----> g(y) = y² − 8

Area = ∫₋₈⁵ (32 − 3y) − (y² − 8) dy
Area = ∫₋₈⁵ (40 − 3y − y²) dy
Area = (40y − 3/2 y² − 1/3 y³) |₋₈⁵
Area = 40(5+8) − 3/2 (25−64) − 1/3 (125+512)
Area = 520 + 117/2 − 637/3
Area = 2197/6
Area = 366 ⅙