Double Integrals and Fubini’s Theorem

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Objectives:

1.: Understand how double integrals give volumes.
2.: Know what Fubini’s theorem says and why it works.
3.: Be able to set up and compute double integrals over arbitrary regions in the $xy$ -plane using Cartesian coordinates.

Recap Video

The following video recaps the ideas of the section.

To recap:

Given a function $f(x,y)$ and a region $R$ in the $xy$ -plane, the double integral $\displaystyle \iint _R f(x,y)dA$ represents the signed volume under $f(x,y)$ lying over the region $R$ in the $xy$ -plane.

Fubini’s Theorem The double integral of a continuous $f(x,y)$ over a rectangle $R = [a,b] \times [c,d]$ can be computed as iterated integrals in either order ( $dx dy$ or $dy dx$ ): $\iint _R f(x,y)dA = \int _a^b \int _c^d f(x,y)dydx = \int _c^d \int _a^b f(x,y)dx dy.$

Test your understanding with the following questions.

The double integral $\displaystyle \iint _R f(x,y)dA$ represents what?

The signed surface area of the solid under $f(x,y)$ and over the region $R$ The signed volume of the region under $f(x,y)$ and over the region $R$ The unsigned surface area of the region under $f(x,y)$ and over the region $R$ The unsigned volume of the region under $f(x,y)$ and over the region $R$

To find the area of a two dimensional region $R$ in the $xy$ -plane, we can evaluate $\displaystyle \iint _R \answer {1}dA$ .

Example Video

Here is a video working through an example of setting up double integrals.

Set up the bounds for the double integral of $f(x,y)$ over the region in the $xy$ -plane bounded by $y = \sqrt {x}$ , $y = 1/x$ , and $x=4$ .

Problems

If $R$ is the region $x^2+y^2 \leq 4$ in the $xy$ -plane, then using geometry we can find that $\iint _R 3dA = \answer {12\pi }.$

Double Integrals over Rectangles

Represent $\displaystyle \iint _R x^3ydA$ , where $R = [-1,1] \times [0,2]$ , as an integral in the order $dy\,dx$ and $dx\,dy$ , and then evaluate the integral.

In the order $dydx$ , we get (fill in the bounds) $\iint _R x^3ydA = \int _{\answer {-1}}^{\answer {1}} \int _{\answer {0}}^{\answer {2}} x^3ydy\,dx.$
In the order $dx\,dy$ , we get (fill in the bounds) $\iint _R x^3ydA = \int _{\answer {0}}^{\answer {2}} \int _{\answer {-1}}^{\answer {1}}x^3ydx\,dy.$
Evaluate in either order to get $\iint _R x^3ydA = \answer {0}.$

Evaluate the integral $\displaystyle \iint _R ye^{-xy}dA$ , where $R = [0,2] \times [0,3]$ .

To integrate with respect to $y$ first would require integration by parts (which we try to avoid if possible), so instead we will use the order $dx\,dy$ . In this case, we get (fill in all steps):

$\begin{eqnarray*} \iint_R ye^{-xy}dA &=& \int_{y=\answer{0}}^{y = \answer{3}} \int_{x = \answer{0}}^{x = \answer{2}} ye^{-xy}dx\,dy \\ &=& \int_{y=0}^{y=3} \left(\answer{-e^{-xy}} \right)\Big|_{x=0}^{x=2} dy \\ &=& \int_{y=0}^{y=3} (\answer{-e^{-2y}+1})dy \\ &=& \left(\answer{\frac{1}{2}e^{-2y}+y} \right)\Big|_{y=0}^{y=3} \\ &=& \answer{\frac{1}{2}e^{-6}+\frac{5}{2}} \end{eqnarray*}$

Double Integrals over General Regions

Method: To determine the bounds for the iterated integrals over regions that aren’t rectangles, look at cross sections.

Set up the double integral $\displaystyle \iint _D xdA$ , where $D$ is the region bounded by $x=0$ , $y = 0$ , $2x+y=2$ , in the order $dy\,dx$ , and then $dx\,dy$ , and then evaluate the integral.

In the order $dy\,dx$ , the iterated integral bounds are: $\int _{\answer {0}}^{\answer {1}} \int _{\answer {0}}^{\answer {2-2x}} x dy\,dx.$
In the order $dx\,dy$ , the iterated integral bounds are: $\int _{\answer {0}}^{\answer {2}} \int _{\answer {0}}^{\answer {1-\frac {y}{2}}} x dx\,dy.$
The integral evaluates to $\iint _D xdA = \answer {1/3}$

Use a double integral to calculate the area of the region $R$ bounded by the curves $y^2=2x$ and $y=x$ in the $xy$ -plane.

To find the area of $R$ , we want to integrate the function $f(x,y) = \answer {1}$ over $R$ . When we integrate in the order $dx\,dy$ , the bounds and integrand will be: $\int _{\answer {0}}^{\answer {2}} \int _{\answer {y^2/2}}^{\answer {y}} \answer {1} dx\,dy.$ When we evaluate the integral, we get an area of $\answer {2/3}$

Set up the integral of a function $f(x,y)$ over the region $R$ bounded by $y=x$ , $y=-x$ , and $y=1$ in the order $dy\,dx$ and $dx\,dy$ .

In the order $dx\,dy$ , the integral is $\int _{\answer {0}}^{\answer {1}} \int _{\answer {-y}}^{\answer {y}} f(x,y) dx\,dy.$
In the order $dy\,dx$ , the integral needs to be split up, and it is: $\int _{-1}^{\answer {0}} \int _{\answer {-x}}^{\answer {1}} f(x,y)dy\,dx + \int _{\answer {0}}^{\answer {1}} \int _{\answer {x}}^{\answer {1}} f(x,y) dy\,dx.$

Evaluate the integral $\displaystyle \int _0^3 \int _{y^2}^9 y\cos (x^2)dx\,dy$ .

We will need to switch the order of integration, so first provide a sketch of the region. Then switching the bounds gives $\int _0^3 \int _{y^2}^9 y\cos (x^2)dx\,dy = \int _{\answer {0}}^{\answer {9}} \int _{\answer {0}}^{\answer {\sqrt {x}}}y\cos (x^2)dy\,dx.$ Evaluating this integral gives an answer of $\answer {\sin (81)/4}$

Use a double integral to find the volume of the tetrahedron bounded by the planes $x+y+z=1$ , $x=0$ , $y =0$ , and $z=0$ .

This volume can be represented as the volume of the function $f(x,y) = \answer {1-x-y}$ over the region in the $xy$ -plane bounded by the lines $y=0$ , $x=0$ , and $y = \answer {1-x}$ . The volume can be represented as an integral $\int _{\answer {0}}^{\answer {1}}\int _{\answer {0}}^{\answer {1-x}} \answer {1-x-y} dy\,dx.$ This integral evaluates to an answer of $\answer {1/6}$