3.1: Double Integrals

In single-variable calculus, differentiation and integration are thought of as inverse operations. For instance, to integrate a function $f (x)$ it is necessary to find the antiderivative of $f$, that is, another function $F(x)$ whose derivative is $f (x)$. Is there a similar way of defining integration of real-valued functions of two or more variables? The answer is yes, as we will see shortly. Recall also that the definite integral of a nonnegative function $f (x) \ge 0$ represented the area “under” the curve $y = f (x)$. As we will now see, the double integral of a nonnegative real-valued function $f (x, y) \ge 0$ represents the volume “under” the surface $z = f (x, y)$.

Let $f (x, y)$ be a continuous function such that $f (x, y) \ge 0 \text{ for all }(x, y)$ on the rectangle $R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}$ in $\mathbb{R}^2$. We will often write this as $R = [a,b] \times [c,d]$. For any number $x∗$ in the interval $[a,b]$, slice the surface $z = f (x, y)$ with the plane $x = x∗$ parallel to the $yz$-plane. Then the trace of the surface in that plane is the curve $f (x∗, y)$, where $x∗$ is fixed and only $y$ varies. The area $A$ under that curve (i.e. the area of the region between the curve and the $x y$-plane) as $y$ varies over the interval $[c,d]$ then depends only on the value of $x∗$. So using the variable $x$ instead of $x∗$, let $A(x)$ be that area (see Figure 3.1.1).

Figure 3.1.1 The area $A(x) \text{ varies with }x$

Then $A(x) = \int_c^d f (x, y)d y$ since we are treating $x$ as fixed, and only $y$ varies. This makes sense since for a fixed $x$ the function $f (x, y)$ is a continuous function of $y$ over the interval $[c,d]$, so we know that the area under the curve is the definite integral. The area $A(x)$ is a function of $x$, so by the “slice” or cross-section method from single-variable calculus we know that the volume $V$ of the solid under the surface $z = f (x, y)$ but above the $x y$-plane over the rectangle $R$ is the integral over $[a,b]$ of that cross-sectional area $A(x)$:

$$V = \int_a^b A(x)dx = \int_a^b \left [\int_c^d f (x, y)d y \right ] dx \label{Eq3.1}$$

We will always refer to this volume as “the volume under the surface”. The above expression uses what are called iterated integrals. First the function $f (x, y)$ is integrated as a function of $y$, treating the variable $x$ as a constant (this is called integrating with respect to \(y\)). That is what occurs in the “inner” integral between the square brackets in Equation $\ref{Eq3.1}$. This is the first iterated integral. Once that integration is performed, the result is then an expression involving only $x$, which can then be integrated with respect to $x$. That is what occurs in the “outer” integral above (the second iterated integral). The final result is then a number (the volume). This process of going through two iterations of integrals is called double integration, and the last expression in Equation $\ref{Eq3.1}$ is called a double integral.

Notice that integrating $f (x, y)$ with respect to $y$ is the inverse operation of taking the partial derivative of $f (x, y)$ with respect to $y$. Also, we could just as easily have taken the area of cross-sections under the surface which were parallel to the $xz$-plane, which would then depend only on the variable $y$, so that the volume $V$ would be

$$V = \int_c^d \left [\int_a^b f (x, y)dx \right ] dy \label{Eq3.2}$$

It turns out that in general the order of the iterated integrals does not matter. Also, we will usually discard the brackets and simply write

$$V=\int_c^d \int_a^b f (x, y)dx d y \label{Eq3.3}$$

where it is understood that the fact that $dx$ is written before $d y$ means that the function $f (x, y)$ is first integrated with respect to $x$ using the “inner” limits of integration $a \text{ and }b$, and then the resulting function is integrated with respect to y using the “outer” limits of integration $c \text{ and }d$. This order of integration can be changed if it is more convenient.

Example 3.1

Find the volume $V$ under the plane $z = 8x +6y$ over the rectangle $R = [0,1]\times [0,2]$.

Solution

We see that $f (x, y) = 8x+6y \ge 0 \text{ for }0 \le x \le 1 \text{ and }0 \le y \le 2$, so:

$$\nonumber \begin{align} V&=\int_0^2 \int_0^1 (8x+6y)dx \,d y \\[4pt] \nonumber &= \int_0^2 \left ( 4x^2 +6x y \big |_{x=0}^{x=1} \right )dy \\[4pt] \nonumber &=\int_0^2 (4+6y)d y \\[4pt] \nonumber &=4y+3y^2 \big |_0^2 \\[4pt] &=20 \end{align}$$

Suppose we had switched the order of integration. We can verify that we still get the same answer:

$$\nonumber \begin{align}V&=\int_0^1 \int_0^2 (8x+6y)dy \,d x \\[4pt] \nonumber &=\int_0^1 \left ( 8x y+3y^2 \big |_{y=0}^{y=2} \right )dx \\[4pt] \nonumber &=\int_0^1 (16x+12)dx \\[4pt] &=8x^2 +12x \big |_0^1 \\[4pt] \nonumber &=20 \end{align}$$

Example 3.2

Find the volume $V$ under the surface $z = e^{x+y}$ over the rectangle $R = [2,3] \times [1,2]$.

Solution

We know that $f (x, y) = e^{x+y} > 0 \text{ for all }(x, y)$, so

$$\nonumber \begin{align} V &=\int_1^2 \int_2^3 e^{x+y} dx\, d y \\[4pt] \nonumber &= \int_1^2 \left (e^{x+y} \big |_{x=2}^{x=3} \right )dy \\[4pt] \nonumber &= \int_1^2 (e^{y+3}-e^{y+2})dy \\[4pt] \nonumber &= e^{y+3}-e^{y+2} \big |_1^2 \\[4pt] \nonumber &= e^5 − e^4 −(e^4 − e^3 ) = e^5 −2e^4 + e^3 \end{align}$$

Recall that for a general function $f (x)$, the integral $\int_a^b f (x)dx$ represents the difference of the area below the curve $y = f (x)$ but above the $x$-axis when $f (x) \ge 0$, and the area above the curve but below the $x$-axis when $f (x) \le 0$. Similarly, the double integral of any continuous function $f (x, y)$ represents the difference of the volume below the surface $z = f (x, y)$ but above the $x y$-plane when $f (x, y) \ge 0$, and the volume above the surface but below the $x y$-plane when $f (x, y) \le 0$. Thus, our method of double integration by means of iterated integrals can be used to evaluate the double integral of any continuous function over a rectangle, regardless of whether $f (x, y) \ge 0$ or not.

Example 3.3

Evaluate $\int_0^{2\pi}\int_0^{\pi}sin(x+y)dx\,dy$

Solution

Note that $f (x, y) = sin(x+ y)$ is both positive and negative over the rectangle $[0,\pi]\times [0,2\pi]$. We can still evaluate the double integral:

$$\nonumber \begin{align} \int_0^{2\pi} \int_0^{\pi}sin(x+y)dx\,dy &=\int_0^{2\pi} \left (-cos(x+y)\big |_{x=0}^{x=\pi} \right )dy \\[4pt] \nonumber &= \int_0^{2\pi} (-cos(y+\pi)+cos{\,y})dy \\[4pt] \nonumber &=-sin(y+\pi)+sin{\,y}\big |_0^{2\pi} = -sin{\,3\pi}+sin{\,2\pi} - (-sin{\,\pi}+sin{\,0}) \\[4pt] \nonumber &=0 \end{align}$$