Double Integrals and the Volume Under a Surface

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

The Volume Under a Surface

Consider the rectangle given by $$R=[a,b]\times[c,d]=\{(x,y)\mid a\leq x\leq b, c\leq x\leq d\}. $$ We wish to construct a (regular) partition of $R.$ To do so, let $$ a=x_0 < x_1 < x_2 < \cdots < x_{i-1} < x_i < \cdots < x_n=b $$ be a partition of $[a,b]$ into subintervals each of width $\Delta x=(b-a)/n.$ Similarly let $$c=y_0 < y_1 < y_2 < \cdots < y_{j-1} < y_j < \cdots < y_m=d$$ be a partition of $[c,d]$ into subintervals each of width $\Delta y=(d-c)/m.$ The vertical lines $x=x_i$ for $0\leq i \leq n$ and the horizontal lines $y=y_j$ for $0\leq j \leq m$ form subrectangles which partition $R$ as shown below. Since each of the subrectangles have the same area, namely $\Delta A=\Delta x \, \Delta y$ we call this type of partition $\mathcal{P}$ a regular partition of the rectangle $R.$

partition-pics
A regular partition of $[a,b]\times[c,d]$ with subrectangles $R_{ij}$ with $1\leq i \leq n,$ $1\leq j \leq m.$
partition-xy-pics
The regular partition of $[-1,2] \times [-2,2]$ with $n=6$ and $m=8.$

For example, let $R=[-1,2]\times[-2,2].$ We have sketched the graph of the regular partition of $R$ with $n=6$ subintervals of $[-1,2]$ and $m=8$ subintervals of $[-2,2].$ For this partition we have $\Delta x=1/2$ and $\Delta y=1/2,$ making the area for each subrectangle $\Delta A =1/4.$

If we increase the number of subintervals of both $[a,b]$ and $[c,d],$ then the number of subrectangles of $R$ increases. This process is illustrated below.

Now we assume that we have a function $f$ of two variables whose domain contains a rectangle $R$ and that we have a regular partition $\mathcal{P}$ of $R$ as described above. From each subrectangle in $\mathcal{P}$ we choose a representative point $\left(x_{ij}^*,y_{ij}^*\right)$ and form the sum, \begin{equation} \label{dbrmsum} \sum_{i=1}^n \sum_{j=1}^m f\left(x_{ij}^*,y_{ij}^*\right)\Delta A \end{equation} where $\Delta A$ is the area of each subrectangle. A summation of this form is called a Riemann sum of $f$ with respect to the partition $\mathcal{P}$ and the subrectangle representatives $(x_{ij}^*,y_{ij}^*).$ Notice that a regular partition of $R$ is completely determined by $n$ and $m.$

The next three examples are an illustration of how Riemann sums can be used to estimate the volume of the 3-dimensional solid region over the $xy$-plane bound below by $R$ and above by the surface $z=f(x,y)$.

partition-xy-5-pics
A regular partition of $R$ with $n=2, m=3.$
R-sum-1-pics

Example. Consider the surface and the rectangle \begin{equation} z=5 – \frac{1}{4} x^2 – \frac{1}{5} y^2 \qquad R=\{(x,y)\mid 0\leq x \leq 2, 0\leq y\leq 3\}\end{equation} Using the lower left corners as subrectangle representatives, determine the Riemann sums when $n=2$ and $m=3.$

R-sum-3-surface-pics
Approximating the volume between the surface $z=5 – (1/4)x^2 – (1/5)y^2$ and the $xy$-plane over the rectangular region $R=[0,2]\times [0,3].$

Solution. When $n=2$ and $m=3$ we have $$ \Delta x=\frac{2-0}{2}=1, \qquad \Delta y=\frac{3-0}{2}=1, \qquad \text{and} \qquad \Delta A=\Delta x \Delta y =1. $$ We are choosing the lower left corners as subrectangle representatives (using $0<1<2$ and $0<1<2<3$), that is, \begin{align} \left(x_{11}^*,y_{11}^*\right)=(0,0) \qquad \left(x_{12}^*,y_{12}^*\right)=(0,1) \qquad \left(x_{13}^*,y_{13}^*\right)=(0,2) \\ \left(x_{21}^*,y_{21}^*\right)=(1,0) \qquad \left(x_{22}^*,y_{22}^*\right)=(1,1) \qquad \left(x_{24}^*,y_{23}^*\right)=(1,2) \end{align} as shown. Now we find the Riemann sum for this partition and chosen representatives: \begin{align*} & \sum_{i=1}^2 \sum_{j=1}^3 f\left(x_{ij}^*, y_{ij}^*\right)\Delta A \sum_{i=1}^2 \sum_{j=1}^3 f\left(x_{ij}^*,y_{ij}^*\right) \\ & = \sum_{i=1}^2 \left[ f\left(x{i1}^*,y_{i1}^*\right) + f\left(x_{i2}^*,y_{i2}^*\right)+ f\left(x_{i3}^*,y_{i3}^*\right)\right] \\ & = f\left(x_{11}^*,y_{11}^*\right) + f\left(x_{12}^*,y_{12}^*\right)+ f\left(x_{13}^*,y_{13}^*\right) + f\left(x_{21}^*,y_{21}^*\right) + f\left(x_{22}^*,y_{22}^*\right) + f\left(x_{23}^*,y_{23}^*\right) \\ & = f(0,0) + f(0,1)+ f(0,2)+f(1,0) + f(1,1) + f(1,2) \\ & = 5+\frac{24}{5}+\frac{21}{5}+\frac{19}{4}+\frac{91}{20}+\frac{79}{20} = \frac{109}{4}=27.25. \end{align*}

partition-xy-6-pics
A regular partition of $R$ with $n=4, m=6.$
R-sum-2-pics

Example. Consider the surface and the rectangle \begin{equation} z=5 – \frac{1}{4} x^2 – \frac{1}{5} y^2 \qquad R=\{(x,y)\mid 0\leq x \leq 2, 0\leq y\leq 3\} \end{equation} Using the lower left corners as subrectangle representatives, determine the Riemann sums when $n=4$ and $m=6.$

Solution. When $n=4$ and $m=6$ we have $$ \Delta x=\frac{2-0}{4}=\frac{1}{2}, \qquad \Delta y=\frac{3-0}{6}=\frac{1}{2}, \qquad \text{and} \qquad \Delta A=\Delta x \Delta y =\frac{1}{4} $$ Again using lower left corners as subrectangle representatives, we find the Riemann sum for this partition to be:

\begin{align*} & \sum _{i=1}^4 \sum _{j=1}^6 f\left(x_{ij}^*,y_{ij}^*\right)\Delta A \\ & = \frac{1}{4}\left[\sum _{i=1}^4 \sum _{j=1}^6 f\left(x_{ij}^*,y_{ij}^*\right)\right] \\ & \qquad = \frac{1}{4} \left[ f\left(0,0\right) + f\left(0,\frac{1}{2}\right) + f\left(0,1\right) + f\left(0,\frac{3}{2}\right) \right. \\ &\qquad\qquad \left. + \, f\left(0,2\right) + \, f\left(0,\frac{5}{2}\right) + \, f\left(\frac{1}{2},0\right) + \, f\left(\frac{1}{2},\frac{1}{2}\right) \right. \\ &\qquad\qquad \left. + \, f\left(\frac{1}{2},1\right) + \, f\left(\frac{1}{2},\frac{3}{2}\right) + \, f\left(\frac{1}{2},2\right) + \, f\left(\frac{1}{2},\frac{5}{2}\right) \right. \\ &\qquad\qquad \left. + \, f\left(0,1\right) + \, f\left(1,\frac{1}{2}\right) + \, f\left(1,1\right) + \, f\left(1,\frac{3}{2}\right) \right. \\ &\qquad\qquad \left. + \, f\left(1,2\right) + \, f\left(1,\frac{5}{2}\right) + \, f\left(\frac{3}{2},0\right) + \, f\left(\frac{3}{2},\frac{1}{2}\right) \right. \\ & \qquad\qquad \left. + \, f\left(\frac{3}{2},1\right) + \, f\left(\frac{3}{2},\frac{3}{2}\right) + \, f\left(\frac{3}{2},2\right) + \, f\left(\frac{3}{2},\frac{5}{2}\right) \right] \\ & \qquad = \left(\frac{1}{4}\right)\frac{415}{4}=25.9375. \end{align*}
partition-xy-7-pics
Double Integrals and the Volume Under a Surface

Example. Consider the surface and the rectangle \begin{equation}z = 5 – \frac{1}{4} x^2 – \frac{1}{5} y^2 \qquad R=\{(x,y)\mid 0\leq x \leq 2, 0\leq y\leq 3\}\end{equation} Using the lower left corners as subrectangle representatives, determine the Riemann sums when $n=6$ and $m=12.$

Solution. When $n=6$ and $m=12$ we have $$ \Delta x=\frac{2-0}{6}=\frac{1}{3}, \qquad \Delta y=\frac{3-0}{12}=\frac{1}{4}, \qquad \text{and} \qquad \Delta A=\Delta x \Delta y =\frac{1}{12} $$ Using lower left corners as subrectangle representatives we find the Riemann sum for this partition to be

\begin{align*} & \sum _{i=1}^6 \sum _{j=1}^{12} f\left(x_{ij}^*,y_{ij}^*\right)\Delta A = \frac{1}{12}\left[\sum _{i=1}^6 \sum _{j=1}^{12} f\left(x_{ij}^*,y_{ij}^*\right)\right] \\ &\qquad = \frac{1}{12} \left[ f\left(0,0\right) + \cdots +f\left(0,\frac{11}{4}\right) + f\left(\frac{1}{3},0\right) + \cdots +f\left(\frac{1}{3},\frac{11}{4}\right) \right. \\ & \qquad \qquad\qquad\left. + \cdots + f\left(\frac{4}{3},0\right) + \cdots +f\left(\frac{4}{3},\frac{11}{4}\right) + f\left(\frac{5}{3},0\right) + \cdots +f\left(\frac{5}{3},\frac{11}{4}\right) \right] \\ & \qquad= \frac{1}{12}\left(\frac{18223}{60}\right)=\frac{14923}{60} =25.3097. \end{align*}

As shown below, the volume of the solid region that lies directly above $R$ and below the surface $z=f(x,y)$ is approximately $25.3097.$

Double Integrals and the Volume Under a Surface
Riemann sum approximation for volume is $25.3097$
Double Integrals and the Volume Under a Surface
The three-dimensional solid that lies directly above $R=[0,2]\times [0,3]$ and below the surface $z=5 – (1/4)x^2 – (1/5)y^2.$

Definition. Let $f$ be defined on the rectangle $R$ and suppose that $f(x,y)\geq 0$ on $R.$ Then the volume $V$ of the solid region that lies directly above $R$ and below the surface $z=f(x,y)$ is \begin{equation} \label{defvolurie} V=\lim_{m,n \to \infty} \sum_{i=1}^n \sum_{j=1}^m f(x_{ij}^*,y_{ij}^*) \, \Delta A \end{equation} if this limit exists.

Midpoint Rule

It can be proven that if $f$ is a continuous function on $R,$ then the limit in \eqref{defvolurie} always exists, no matter how the subrectangle representatives $(x_{ij}^*, y_{ij}^*)$ are chosen. For example, if $a=x_0 < x_1 < x_2 < \cdots < x_{i-1} < x_i < \cdots < x_n=b$ is a regular partition of $[a,b]$ and $c=y_0 < y_1 < y_2 < \cdots < y_{j-1} < y_j < \cdots < y_m=d$ is a regular partition of $[c,d],$ then we can use midpoints, namely, from each subrectangle $[x_i,x_{i+1}]\times [y_j, y_{j+1}]$ we choose subrectangle representatives. This is called the midpoint rule.

partition-xy-8-pics
Using midpoints as subrectangle representatives.

Example. Consider the surface and the rectangle \begin{equation} z = 2+\frac{1}{2}x^2+\frac{1}{3}y^2 \qquad R=\{(x,y)\mid -1\leq x \leq 2, -3\leq y\leq 4\}\end{equation} Use the midpoint rule to find the Riemann sum determined by the regular partition of $R$ with $n=12$ and $m=28$ to estimate the volume of the solid that lies under the graph of the surface $z=f(x,y)$ and directly above the rectangle $R.$

Solution. When $n=12$ and $m=28$ we have $$ \Delta x=\frac{2-(-1)}{12}=\frac{1}{4}, \quad \Delta y=\frac{4-(-3)}{28}=\frac{1}{4},
\quad \text{and} \quad \Delta A=\Delta x \Delta y =\frac{1}{16}. $$ The subrectangle representations are: \begin{align} (x_{ij}^*, y_{ij}^*) & = \left( \frac{(-1+\frac{1}{4}(i-1))+(-1+\frac{1}{4} i)}{2}, \frac{-3+\frac{1}{4}(j-1))+(-3+\frac{1}{4} j)}{2} \right) \\ & =\left( -\frac{9}{8}+\frac{1}{4}i, -\frac{25}{8}+\frac{1}{4} j \right) \end{align} Using midpoints we find the Riemann sum for this partition to be:

\begin{align*} & \sum _{i=1}^{12} \sum _{j=1}^{28} f\left(x_{ij}^*,y_{ij}^*\right)\Delta A = \frac{1}{16}\left[\sum _{i=1}^{12} \sum _{j=1}^{28} f\left(-\frac{9}{8}+\frac{1}{4}i, -\frac{25}{8}+\frac{1}{4} j \right)\right] \\ & \qquad = \frac{1}{16} \left[ f\left(-\frac{7}{8},-\frac{23}{8}\right) +f\left(-\frac{7}{8},-\frac{21}{8}\right) + f\left(-\frac{7}{8},-\frac{19}{8}\right) \right. \\ & \qquad \qquad \qquad \qquad \left.+\cdots +f\left(\frac{15}{8},\frac{27}{8}\right) + f\left(\frac{15}{8},\frac{29}{8}\right) +f\left(\frac{15}{8},\frac{31}{8}\right) \right] \\ & \qquad = \frac{1}{16}\left(\frac{10591}{8}\right)=82.7421875. \end{align*}
R-sum-5-surface-pics
Riemann sum approximation for volume is $82.7422$
Double Integrals and the Volume Under a Surface
The three-dimensional solid that lies directly above $R=[-1,2]\times [-3,4]$ and below the surface $z=2+(1/2)x^2+(1/3)y^2 .$

Definition of Double Integral

In our previous examples, we used regular partitions to form Riemann sums for functions defined over rectangular regions. This is not necessity. To measure the size of the rectangles in the partition $P,$ we define the norm $|P|$ of the partition to be the length of the longest diagonal of any of the subrectangles in the partition. Refine a partition $P$ by subdividing the cells in such a way that the norm decreases. When this process is applied to the Riemann sum and the norm decreases to zero we have the double integral of $f$ over $R.$

Definition. If a function $f$ is defined on a closed, bounded rectangular region $R$ in the $x y$-plane, then the double integral of $f$ over $R$ is defined by \begin{equation} \iint_R f(x,y)\,d A=\lim {|P|\to 0}\sum_{k=1}^N f\left(x_k^*, y_k^*\right)\triangle A_k\end{equation} provided this limit exists, in which case, $f$ is said to be integrable over $R.$

Let $f(x,y)$ be a function that is continuous on the region $D$ that can be contained in a rectangle $R.$ Define the function $F(x,y)$ on $R$ as $f(x,y)$ if $(x,y)$ is in $D$ and 0 otherwise. If $F$ is integrable over $R,$ we say that $f$ is integrable over $D,$ and the double integral of $f$ over $D$ is defined as \begin{equation} \label{leftrightint} \iint_D f(x,y)\,d A=\iint_R F(x,y)\,d A. \end{equation}

The function $F(x,y)$ may have discontinuities on the boundary of $D,$ but if $f(x,y)$ is continuous on $D$ and the boundary of $D$ is fairly “well behaved”, then it can be shown that the double integral of the right side of \eqref{leftrightint} exists and hence that the double integral of the left side of \eqref{leftrightint} exists.

Properties of Double Integrals

Theorem. Assume that all the given integrals exist on a rectangular region $R.$

(1) For constants $a$ and $b,$ $$ \iint_R [a f(x,y)+b g(x,y)]\, dA =a \iint_R f(x,y) \, d A+b\iint_R g(x,y) \, dA. $$

(2) If $f(x,y)\geq g(x,y)$ throughout a rectangular region $R,$ then $$
\iint_R f(x,y)d A\geq \iint_R g(x,y)d A. $$

(3) If the rectangular region of integration $R$ is subdivided into two (disjoint) subrectangles $R_1$and $R_2,$ then $$ \iint_R f(x,y)d A= \iint_{R_1} f(x,y)d A+\iint_{R_2} f(x,y)d A. $$

How do we perform calculations with double integrals?

Exercises on Double Integrals

Exercise. If $f$ is a constant function, say $f(x,y)=k$ and $R = [a,b] \times [c,d] $ show that \begin{equation} \iint_R k dA=k(b-a)(d-c). \end{equation}

Exercise. Approximate the given double integral by dividing the rectangle $R$ with vertices $(0,0),$ $(4,0),$ $(4,2),$ and $(0,2)$ into eight squares and find the sum.

$(1) \quad \displaystyle \iint_R (x+y) \, dA$

$(2) \quad \displaystyle \iint_R x y \, dA$

$(3) \quad \displaystyle \iint_R (x^2+y^2) \, dA$

$(4) \quad \displaystyle \iint_R\frac{1}{(x+1)(y+1)} \, dA$

Exercise. Find an approximation for the volume $V$ of the solid lying under the graph of the elliptic paraboloid $z=8-2x^2-y^2$ and above the rectangular region $R=[0,1]\times [0,2].$ Use a regular partition $\mathcal{P}$ of $R$ with $m=n=2,$and choose the subrectangular representation point $(x_{ij}^*,y_{ij}^*)$ as indicated.

(1) The lower left-hand corner of $R_{ij}.$

(2) The upper left-hand corner of $R_{ij}.$

(3) The lower right-hand corner of $R_{ij}.$

(4) The upper right-hand corner of $R_{ij}.$

(5) The center of $R_{ij}.$

Exercise. Calculate the double Riemann sum of $f$ for the partition of $R$ given by the indicated lines and the given choice of $(x_{ij}^*,y_{ij}^*).$ Also calculate the norm of the partition.

$(1) \quad f(x,y)=x^2+4y,$ $R=[0,2]\times[0,3],$ $x=1, y=1, y=2$; $(x_{ij}^*, y_{ij}^*)$ is the upper right corner of $R_{ij}$

$(2) \quad f(x,y)=x^2+4y,$ $R=[0,2]\times[0,3],$ $x=1,$ $y=1,$ $y=2;$ $(x_{ij}^*, y_{ij}^*)$ is the center of $R_{ij}$

$(3) \quad f(x,y)=xy-y^2,$ $R=[0,5]\times[0,4],$ $x=1,$ $x=2,$ $x=3,$ $x=4, y=2$; $(x_{ij}^*, y_{ij}^*)$ is the center of $R_{ij}$

$(4) \quad f(x,y)=2x+x^2y,$ $R=[-2,2]\times[-1,1],$ $x=-1,$ $x=0,$ $x=1,$ $y=-1/2,$ $y=0,$ $y=1/2$; $(x_{ij}^*,y_{ij}^*)$ is the lower left corner of $R_{ij}$

$(5) \quad f(x,y)=x^2-y^2,$ $R=[0,5]\times[0,2],$ $x=1,$ $x=3,$ $x=4,$ $y=1/2,$ $y=1$; $(x_{ij}^*,y_{ij}^*)$ is the upper left corner of $R_{ij}$

$(6) \quad f(x,y)=5xy^2,$ $R=[1,3]\times[1,4],$ $x=1.8,$ $x=2.5,$ $y=2, y=3$; $(x_{ij}^*,y_{ij}^*)$ is the lower right corner of $R_{ij}$

Exercise. Calculate the double Riemann sum of $f$ for the partition of $R$ given by the indicated lines and the given choice of $(x_{ij}^*,y_{ij}^*).$ Also calculate the norm of the partition.

$(1) \quad f(x,y)=x^2+4y,$ $R=[0,2]\times[0,3],$ $x=1, y=1, y=2$; $(x_{ij}^*, y_{ij}^*)$ is the upper right corner of $R_{ij}$

$(2) \quad f(x,y)=x^2+4y,$ $R=[0,2]\times[0,3],$ $x=1, y=1, y=2$; $(x_{ij}^*, y_{ij}^*)$ is the center of $R_{ij}$

$(3) \quad f(x,y)=xy-y^2,$ $R=[0,5]\times[0,4],$ $x=1, x=2, x=3, x=4, y=2$; $(x_{ij}^*, y_{ij}^*)$ is the center of $R_{ij}$

$(4) \quad f(x,y)=2x+x^2y,$ $R=[-2,2]\times[-1,1],$ $x=-1,$ $x=0,$ $x=1,$ $y=-1/2,$ $y=0,$ $y=1/2;$ $(x_{ij}^*,y_{ij}^*)$ is the lower left corner of $R_{ij}$

$(5) \quad f(x,y)=x^2-y^2,$ $R=[0,5]\times[0,2],$ $x=1,$ $x=3,$ $x=4,$ $y=1/2,$ $y=1;$ $(x_{ij}^*,y_{ij}^*)$ is the upper left corner of $R_{ij}$

$(6) \quad f(x,y)=5xy^2,$ $R=[1,3]\times[1,4],$ $x=1.8,$ $x=2.5,$ $y=2,$ $y=3;$ $(x_{ij}^*,y_{ij}^*)$ is the lower right corner of $R_{ij}$

Exercise. Find the Riemann sum

$$\sum_{i=1}^m\sum_{j=1}^n f(x_{ij}^*,y_{ij}^*) \Delta A$$

of $f$ over the region $R$ with respect to the regular partition $\mathcal{P}$ with the indicated values of $m$ and $n.$

$(1) \quad f(x,y)=2x+3y,$ $R=[0,1]\times[0,3],$ $m=2, n=3$; $(x_{ij}^*,y_{ij}^*)$ is the lower left-hand corner of $R_{ij}$

$(2) \quad f(x,y)=x^2-2y,$ $R=[1,5]\times[1,3],$ $m=4, n=2$; $(x_{ij}^*,y_{ij}^*)$ is the upper right-hand corner of $R_{ij}$

$(3) \quad f(x,y)=x^2+2y^2,$ $R=[-1,3]\times[0,4],$ $m=4, n=2$; $(x_{ij}^*,y_{ij}^*)$ is the upper right-hand corner of $R_{ij}$

$(4) \quad f(x,y)=2xy,$ $R=[-1,1]\times[-2,2],$ $m=4, n=4$; $(x_{ij}^*,y_{ij}^*)$ is the center of $R_{ij}$

Exercise. Evaluate the integral $\iint_R (4-2y)\, dA,$ where $R=[0,1]\times [0,1],$ by identifying it as the volume of a solid.

Exercise. The integral $$ \iint_R \sqrt{9-y^2} \, dA, $$ where $R=[0,4]\times[0,2],$ represents the volume of a solid. Sketch the solid.

Exercise. If $f$ is a constant function, $f(x,y)=k,$ and $R=[a,b]\times [c,d],$ show that $$ \iint_R k \, dA =k(b-a)(d-c). $$

Exercise. If $R=[0,1]\times [0,1],$ show that $$ 0\leq \iint_R \sin(x+y) \, dA \leq 1. $$

Exercise. Evaluate $\iint_{R} y e^{xy} \, dA$ where $R=[0,2]\times [0,1].$