How do you estimate the volume of the solid that is beneath the surface defined by z = xy and above the rectangle defined by x in the range 0 to 6 and y in the range 0 to 4?

To estimate the volume of the solid that lies below the surface z = xy and above the rectangle R defined by 0 ≤ x ≤ 6 and 0 ≤ y ≤ 4, we can use double integration.

The volume V can be represented as:

V = ∬R z dA

Here, R is the projection of the solid onto the xy-plane, which is the rectangle defined by the ranges of x and y.

Now, substituting z = xy, we have:

V = ∬R xy dA

Next, we can set up the double integral over the rectangle R. The limits for x will be from 0 to 6, and for y, from 0 to 4:

V = ∫06 ∫04 xy dy dx

Now, let’s evaluate the inner integral:

∫04 xy dy = x ∫04 y dy = x [ (y²/2) ] |04 = x (16/2) = 8x

So now, we have:

V = ∫06 8x dx

Calculating the outer integral:

∫06 8x dx = 8 [ (x²/2) ] |06 = 8 ( (36/2) - 0 ) = 8 × 18 = 144

Thus, the volume of the solid beneath the surface z = xy and above the rectangle R is:

V = 144

This means the estimated volume of the solid is 144 cubic units.