SOLUTION We first observe that \(S\) is the solid that lies under the
surface \(z = 16 - x^2 - 2y^2\) and
above the square \(R = [0, 2] \times [0,
2]\). This solid was considered in Example 1, but we are now in a
position to evaluate the double integral using Fubini’s Theorem.
Therefore \[
V = \iint_R (16 - x^2 - 2y^2) dA = \int_0^2 \int_0^2 (16 - x^2 - 2y^2)
dy dx
\] \[
= \int_0^2 \left[ 16y - x^2y - \frac{2}{3}y^3 \right]_{y=0}^{y=2} dx =
\int_0^2 \left( 32 - 2x^2 - \frac{16}{3} \right) dx
\] \[
= \int_0^2 \left( \frac{80}{3} - 2x^2 \right) dx = \left[ \frac{80}{3}x
- \frac{2}{3}x^3 \right]_0^2 = \frac{160}{3} - \frac{16}{3} =
\frac{144}{3} = 48
\]
In the special case where \(f(x,
y)\) can be factored as the product of a function of \(x\) only and a function of \(y\) only, the double integral of \(f\) can be written in a particularly simple
form. To be specific, suppose that \(f(x, y) =
g(x)h(y)\) and \(R = [a, b] \times [c,
d]\). Then Fubini’s Theorem gives \[
\iint_R f(x, y) dA = \int_c^d \int_a^b g(x)h(y) dx dy = \int_c^d \left[
\int_a^b g(x)h(y) dx \right] dy
\] In the inner integral, \(y\)
is a constant, so \(h(y)\) is a
constant and we can write \[
\int_a^b g(x)h(y) dx = h(y) \int_a^b g(x) dx
\] since \(\int_a^b g(x) dx\) is
a constant. Therefore, in this case the double integral of \(f\) can be written as the product of two
single integrals: \[ \tag{eq:11}
\iint_R g(x)h(y) dA = \left( \int_a^b g(x) dx \right) \left( \int_c^d
h(y) dy \right) \quad \text{where } R = [a, b] \times [c, d]
\]
