In order to calculate the area within two curves f(x) and g(x), I can use a single integral, and my bounds will be the points of intersection where f(x) = g(x).
Given that f(x) is the red function and g(x) is the blue function, g(x) > f(x) over the region of integration. The points of intersection occur at $(x_1,y_1)$ and $(x_2,y_2)$ Thus, the following integral will yield the enclosed area. $\int_{x_1}^{x_2} (g(x)-f(x)) dx$
How can I extend this process to double integrals? Say I wish to find the enclosed volume between two surfaces f(x,y) and g(x,y).
Given f(x,y) is the blue function and g(x,y) is the purple function, g(x,y) > f(X,y) over the region of integration. The points of intersection occur over a set of points $(x_k,y_k,z_k)$ that lie along a curve. Thus, the double integral that will yield the enclosed volume is $\iint_D (g(x,y)-f(x,y)) dxdy$.
What the bounds for my region D?
Area between two functions
Given the set:
$$ D := \left\{ (x,\,y) \in \mathbb{R}^2 : y \ge x^2, \; y \le 1-x^2 \right\}, $$
the measure of $D$ is equal to:
$$ ||D|| := \iint\limits_D 1\,\text{d}x\,\text{d}y = \int\limits_{x^2 \le 1-x^2} \text{d}x \int_{x^2}^{1-x^2} \text{d}y = \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \left(1-2\,x^2\right)\text{d}x = \frac{2\sqrt{2}}{3}\,. $$
Volume between two functions
Given the set:
$$ \Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : z \ge x^2+y^2, \; z \le 1-x^2-y^2 \right\}, $$
the measure of $\Omega$ is equal to:
$$ ||\Omega|| := \iint\limits_{\Omega} 1\,\text{d}x\,\text{d}y\,\text{d}z = \iint\limits_{x^2+y^2 \le 1-x^2-y^2} \text{d}x\,\text{d}y \int_{x^2+y^2}^{1-x^2-y^2} \text{d}z $$
i.e.
$$ ||\Omega|| = \iint\limits_{x^2+y^2\le\left(\frac{\sqrt{2}}{2}\right)^2} \left(1-2\,x^2-2\,y^2\right) \text{d}x\,\text{d}y = \int_0^{2\pi} \text{d}\theta \int_0^{\frac{\sqrt{2}}{2}} \left(1-2\,\rho^2\right)\rho\,\text{d}\rho = \frac{\pi}{4}\,. $$