# 3.1: Double Integrals

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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).

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}

This page titled 3.1: Double Integrals is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Michael Corral via source content that was edited to the style and standards of the LibreTexts platform.