To find the volume of the region under the graph of f(x,y) = 3x + y + 1 and above the region y^2 ≤ x, 0 ≤ x ≤ 9, we need to integrate f(x,y) over the given region.
The region is bounded by the curves y = sqrt(x) and y = -sqrt(x), and the lines x = 0 and x = 9. So, we can set up the double integral as follows:
∫(from 0 to 9)∫(from -sqrt(x) to sqrt(x)) (3x + y + 1) dy dx
Evaluating the inner integral with respect to y, we get:
∫(from 0 to 9) [ (3x + y + 1)y ] (from -sqrt(x) to sqrt(x)) dx
= ∫(from 0 to 9) [ (3x + sqrt(x) + 1)sqrt(x) - (3x - sqrt(x) + 1)sqrt(x) ] dx
= ∫(from 0 to 9) [ 2sqrt(x) + 1 ] dx
= [ (4/3)x^(3/2) + x ] (from 0 to 9)
= 108 + 9sqrt(3)
Therefore, the volume of the region under the graph of f(x,y) = 3x + y + 1 and above the region y^2 ≤ x, 0 ≤ x ≤ 9 is 108 + 9sqrt(3) cubic units.