Volume under a surface

Just like the integral $\int_a^b f(x)dx$ gives the area between y=0 and y=f(x) from x=a to x=b, the integral $\int_R fdA$ gives the total volume of the solid which lies between z=0 and z=f(x,y) with a cross section shaped like R.

What's the volume of the solid shown here? (The region R is shown more clearly to the right.)

Clearly, $\int_R f_1 dA$ is the volume between z=0 and z = f₁(x,y) while $\int_R f_2 dA$ is the volume between z=0 and z = f₂(x,y). Thus, the volume between f₁ and f₂ is the difference of these two integrals. Since f₂ is above f₁, the volume of the solid is

$$\mbox{Volume} \quad = \int_R f_2 dA - \int_R f_1 dA$$

$$= \int_R (f_2 - f_1) dA = \int_0^2 \int_0^y (f_2(x,y) - f_1(x,y)) dx dy, \quad \mbox{or}$$

$$= \int_0^2 \int_x^2 (f_2(x,y) - f_1(x,y)) dy dx.$$