Integrals of Functions of Two Variables

Double Integral

Given a function \(f:\mathbb{R}^2 \to \mathbb{R}\) defined on a region \(R \subset \mathbb{R}^2\), we define the double integral of \(f\) over the region \(R\)

\[\iint_R f(x,y)\;dA\]

as the signed volume of the solid between the \(xy\)-plane and the graph of \(f\) over the region \(R\).

By signed volume we mean that the parts of the solid that are below the \(xy\)-plane are considered negative.

The region \(R\) can be a rectangle:

or some other shape. There are some requirements on the region, namely that the boundary of the region is a continuous piecewise smooth simple closed curve: the boundary is a curve that can be parametrized by a vector function \(\mathbf{r}(t)\), \(a \le t \le b\) such that

\(\mathbf{r}(a) = \mathbf{r}(b)\)
\(\mathbf{r}(t_0) \neq \mathbf{r}(t_1)\) for any \(t_0 \in [a,b)\), \(t_1 \in [a,b)\), \(t_0 \neq t_1\).
\(\mathbf{r}\) is continuous on \([a,b]\).
\(\mathbf{r}\) has a continuous and bounded derivative except at at most finitely many points on \([a,b]\).

A double integral of a function \(f\) over such region will be the signed volume of a solid like this:

Formally a double Riemann integral can be defined as a limit of a two-dimensional Riemann sum, similar way as a single variable Riemann integral is defined.

Intuitively we can think about the “quantity” \(dA\) as an infinitesimal area of a single point (you can think about it as an infinitely small rectangle with sides \(dx\) and \(dy\), then \(dA = dx\;dy\)). Then \(f(x,y)\;dA\) represents a “volume” of an infinitely thin rectangular prism whose base has area \(dA\) and whose height is \(f(x,y)\). We then “add” these “volumes” for every point in the region \(R\), representing the (two dimensional) “summation” over all the points in \(R\) by the “double S” sign (for sum, or die Summe in German), getting the integral notation

\[\iint_R f(x,y)\;dA\]

Partial Integrals

Given a function \(f:\mathbb{R}^2 \to \mathbb{R}\) and two real numbers \(a_0\) and \(a_1\), we can define a partial integral of \(f\) with respect to \(x\) from \(a_0\) to \(a_1\):

\[\int_{a_0}^{a_1} f(x,y)\; dx\]

The integration variable is \(x\), and \(y\) is considered to be a constant parameter. As a result, we get a function of \(y\). It will be the signed area of a region in the plane parallel to the \(xz\)-plane, passing through the point \((0,y,0)\), between the \(xy\)-plane and the graph of \(f\), from \(x=a_0\) to \(x = a_1\):

An example: Calculate the partial integral \[\int_0^1 x^2y + xy^3\;dx\]

Another example: Calculate the partial integral \[\int_{y^2}^y x^2 + xy - y^3\;dx\]

Similarly, we can define a partial integral of \(f\) with respect to \(y\) from \(c_0\) to \(c_1\):

\[\int_{c_0}^{c_1} f(x,y)\; dy\]

This time the integration variable is \(y\), and \(x\) is considered to be a constant parameter. As a result, we get a function of \(x\). It will be the signed area of a region in the plane parallel to the \(yz\)-plane, passing through the point \((x,0,0)\), between the \(xy\)-plane and the graph of \(f\), from \(y=c_0\) to \(y = c_1\):

Yet another example: Calculate the partial integral \[\int_0^x x^2y + xy^3\;dy\]

Iterated integrals

Rectangular regions

Suppose we have a double integral over a rectangular region \(R = \{(x,y) \in \mathbb{R}^2 \mid a \le x \le b \text{ and } c \le y \le d\}\). We can write two iterated integrals:

We can calculate the partial integral with respect to \(x\) from \(a\) to \(b\), and then integrate the resulting function of \(y\) from \(c\) to \(d\):

\[\int_c^d \int_a^b f(x,y)\;dx\;dy\]

The following illustration shows the solid corresponding to the double integral, and the region corresponding to the partial integral with respect to \(x\) for one of the \(y\) values between \(c\) and \(d\):
We can calculate the partial integral with respect to \(y\) from \(c\) to \(d\), and then integrate the resulting function of \(x\) from \(a\) to \(b\):

\[\int_a^b \int_c^d f(x,y)\;dy\;dx\]

The following illustration shows the solid corresponding to the double integral, and the region corresponding to the partial integral with respect to \(y\) for one of the \(x\) values between \(a\) and \(b\):

Non-rectangular regions

Things can get more complicated with a non-rectangular region \(R\). The problem is that the limits of the partial integral will also depend on the free variable:

We can calculate the partial integral with respect to \(x\) from \(a(y)\) to \(b(y)\), and then integrate the resulting function of \(y\) from \(c\) to \(d\):

\[\int_c^d \int_{a(y)}^{b(y)} f(x,y)\;dx\;dy\]

This shows one of the slices in the for a specific value of \(y\) in the region \(R\):

The following illustration shows the solid corresponding to the double integral, and the region corresponding to the iterated integral with respect to \(x\) for one of the \(y\) values between \(c\) and \(d\):
We can calculate the partial integral with respect to \(y\) from \(c(x)\) to \(d(x)\), and then integrate the resulting function of \(x\) from \(a\) to \(b\):

\[\int_a^b \int_{c(x)}^{d(x)} f(x,y)\;dy\;dx\]

This shows one of the slices in the for a specific value of \(x\) in the region \(R\):

The following illustration shows the solid corresponding to the double integral, and the region corresponding to the iterated integral with respect to \(y\) for one of the \(x\) values between \(a\) and \(b\):

Intuitively we can think about an iterated integral like \[\int_c^d \int_{a(y)}^{b(y)} f(x,y)\;dx\;dy\] in the following way:

For any given \(y\) such that at least one point \((x,y)\) is in \(R\), the partial integral \[\int_{a(y)}^{b(y)} f(x,y)\;dx\] represents the area of a face that we get by slicing our solid vertically at \(y\), in the direction perpendicular to the \(y\)-axis.. Multiplying this area by an infinitesimal thickness \(dy\), we get \[\left(\int_{a(y)}^{b(y)} f(x,y)\;dx\right)\;dy\] which represents the “volume” of an infinitely thin slice taken from the solid at the given \(y\).

We then take all these (inifinitely many) slices and add their volumes together: \[\int_c^d \left(\int_{a(y)}^{b(y)} f(x,y)\;dx\right)\;dy\]

In a similar way we can interpret the integral \[\int_a^b \int_{c(x)}^{d(x)} f(x,y)\;dy\;dx\] as the sum of “volumes” of infinitely thin “slices” that are perpendicular to the \(x\)-axis: \[\int_a^b\left(\int_{c(x)}^{d(x)} f(x,y)\;dy\right)\;dx\]

Fubini’s Theorem

Now we have three different way how to calculate a volume of a solid like the ones in the examples above:

We can “slice” the solid into infinitely many infinitely thin vertical “french fries”, calculate the “volume” of each “fry”, and add them all together. This will give us a double integral.
We can slice the solid into infinitely many infinitely thin vertical “salami slices” that are perpendicular to the \(y\)-axis, calculate the “volume” of each “slice”, and add them all together. This will give us a “\(dx\;dy\) iterated integral.
We can slice the solid into infinitely many infinitely thin vertical “salami slices” that are perpendicular to the \(x\)-axis, calculate the “volume” of each “slice”, and add them all together. This will give us a “\(dy\;dx\) iterated integral.

Since we are slicing the same solid, we would expect that these three processes will give us the same result. That is indeed the case, provided the function \(f\) and the region \(R\) are nice enough.

Theorem: Let \(f\) be a function of two variables defined on a region \(R\) such that \(f\) and \(R\) are nice¹ Then the double integral of \(f\) over the region \(R\) and the two iterated integrals over the same region are equal:

¹ For some sufficient conditions, look up Fubini’s theorem in your textbook.

\[\iint_R f(x,y)\;dA = \int_c^d \int_{a(y)}^{b(y)} f(x,y)\;dx\;dy = \int_a^b \int_{c(x)}^{d(x)} f(x,y)\;dy\;dx\]

Examples

Integrals over rectangular regions

First we will calculate \[\iint_R xy^2 - x^2 + y\; dA\] where \(R\) is the rectangle with corners \((0,0)\), \((2,0)\), \((2,1)\) and \((0,1)\):

In the next example we will calculate \[\iint_R y^2\cos(x)\; dA\] where \(R\) is the rectangle given by inequalities \(0 \le x \le \pi/2\) and \(1 \le y \le 2\):

Integrals over general regions

First we will calculate \[\iint_R xy^2 - x^2 + y\; dA\] where \(R\) is the triangle with corners \((0,0)\), \((2,0)\), and \((0,1)\):

Next we will calculate the same integral, but slicing horizontally instead of vertically.

As the last example, we will calculate \[\iint_R xy\; \dA\] where \(R\) is the region bounded by the curves \(x = y^2\) and \(x - 2y = 3\).

Note that some regions can be easily sliced in only one of the two ways, and some of them cannot be easily sliced in either of the two ways, and must instead be divided into several smaller regions.