Can't figure out this volume with a double integral question from my textbook.

Question

The solid lies under the hyperboloid z = xy and above the triangle in the xy-plane with vertices (1, 2),(1, 4), and (5, 2). Find the volume of the given solid.

I got that the answer is 42, but the correct answer is 24. Could someone show me how to make the double integral that gets this answer?

Answer 1

V = ∫∫_Rxydxdy, where R is given triangle: 1 ≤ x ≤5; 2 ≤ y ≤ -1/2x + 9/2

V = ∫₁⁵xdx∫₂^-1/2x+9/2ydy = ∫₁⁵xy²/2₂^-1/2x+9/2dx = ∫₁⁵x/2(1/4x² - 9/2x + 81/4 - 4)dx =

(x⁴/32 - 3/4x³ + 65/16x²)₁⁵ = 625/32 - 375/4 + 65·25/16 - 1/32 + 3/4 - 65/16 = 24