find the volume of the solid enclosed by the surface z=x-xy^2+12 and the planes z=0,x=0,x=3 and y=+-2

To find the volume of the solid enclosed by the given surface and planes, we can use the method of double integration.

The volume V of the solid is given by the double integral of the function z(x, y) over the region R in the xy-plane that is bounded by the given planes.

The region R is a rectangle with corners (0, -2), (0, 2), (3, -2), and (3, 2).

So, we can set up the double integral as follows:

V = ∫ (from y=-2 to y=2) ∫ (from x=0 to x=3) (x - xy^2 + 12) dx dy

First, we integrate with respect to x:

= ∫ (from y=-2 to y=2) [0.5x^2 - 0.5x^2*y^2 + 12x] (from x=0 to x=3) dy

= ∫ (from y=-2 to y=2) [0.53^2 - 0.53^2y^2 + 123] dy

= ∫ (from y=-2 to y=2) [4.5 - 4.5y^2 + 36] dy

= ∫ (from y=-2 to y=2) [40.5 - 4.5y^2] dy

Next, we integrate with respect to y:

= [40.5y - 1.5y^3] (from y=-2 to y=2)

= [40.52 - 1.52^3] - [40.5*(-2) - 1.5*(-2)^3]

= [81 - 24] - [-81 - 24]

= 57 - (-105)

= 162

So, the volume of the solid is 162 cubic units.