Wize University Calculus 3 Textbook > Multiple Integrals

Double Integrals over General Regions

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If the region of general shape is defined as D={(x,y)∣a≤x≤b, g1(x)≤y≤g2(x)}, D=\{(x,y)|a\le x\le b,\ g_1(x)\le y\le g_2(x)\},~D={(x,y)∣a≤x≤b, g1​(x)≤y≤g2​(x)}, then the double integral over D can be expressed as:
∬Df(x,y)dA=∫ab∫g1(x)g2(x)f(x,y)dydx\color{blue}\iint_Df(x,y)dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)dydx∬D​f(x,y)dA=∫ab​∫g1​(x)g2​(x)​f(x,y)dydx

Similarly, for a general region defined as D={(x,y)∣c≤y≤d, h1(y)≤x≤h2(y)}, D = \{(x,y)| c \le y \le d,\ h_1(y) \le x \le h_2(y)\},~D={(x,y)∣c≤y≤d, h1​(y)≤x≤h2​(y)}, the double integral over D will be expressed as:
∬Df(x,y)dA=∫cd∫h1(y)h2(y)f(x,y)dxdy\color{blue}\iint_Df(x,y)dA=\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)dxdy∬D​f(x,y)dA=∫cd​∫h1​(y)h2​(y)​f(x,y)dxdy
Remember: Switching the order of integration can make it easier to solve! The important task in changing the order of integrals is determining the limits of integration : )

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Evaluate ∫∫DxeydA\displaystyle\int_{ }^{ }\int_D^{ }xe^ydA∫​∫D​xeydA, where D is the region between y=xy=xy=x and y=x2y=x^2y=x2.

I=∫01∫x2xxey dydx    =∫01x dx∫x2xey dy   =∫01x dx(ey∣x2x)            =∫01x dx ⋅ [ex − ex2]                 =∫01xex dx − ∫01xex2 dx                      A                     BI=\displaystyle\int_0^1\int_{x^2}^{x}xe^y~dydx\newline{}\newline{}
~~~~
=\int_0^1x~dx\int_{x^2}^{x}e^y~dy\newline{}\newline{}
~~~
=\int_0^1x~dx\Bigg(e^y\Bigg|_{x^2}^x\Bigg)\newline{}\newline{}
~~~~~~~~~~~~
=\int_0^1x~dx~\cdot~\Big[e^{x}~-~e^{x^2}\Big]\newline{}\newline{}
~~~~~~~~~~~~~~~~~
=\int_0^1xe^x~dx~-~\int_0^1xe^{x^2}~dx\newline{}

~~~~~~~~~~~~~~~~~~~~~~{\scriptsize\color{magenta}A}~~~~~~~~~~~~~~~~~~~~~ {\scriptsize\color{magenta}B}I=∫01​∫x2x​xey dydx    =∫01​x dx∫x2x​ey dy   =∫01​x dx(ey​x2x​)            =∫01​x dx ⋅ [ex − ex2]                 =∫01​xex dx − ∫01​xex2 dx                      A                     B
Let A \color{magenta}A~A be integrated by parts and let B \color{magenta}B~B be integrated with substitution:
A‾:B‾:∫01xex dx = xex∣01 − ∫01ex dx        ∫01xex2 dx =                      Let u=x2. Then, du2=xdx=e − (ex)∣01= 12∫01eu du                    =e − (e − 1)=12(ex2)∣01                     =1                 =12(e−1)                      \begin{array}{c|c}
{\color{magenta}\underline{A}:} & {\color{magenta}\underline{B}:}
\\\\
\displaystyle\int_0^1xe^x~dx~=~xe^x\Big|_0^1~-~\int_0^1e^x~dx~~~~~~&~~\displaystyle\int_0^1xe^{x^2}~dx~=~~~~~~~~~~~~~~~~~~~~~~{\scriptsize\color{magenta}Let~u=x^2.~Then,~\frac{du}{2}=xdx}\\\\

=e~-~(e^x)\bigg|_0^1 &=~\frac{1}{2}\displaystyle\int_0^1e^u~du~~~~~~~~~~~~~~~~~\\\\~~~=e~-~(e~-~1)&=\frac{1}{2}\Big(e^{x^2}
\Big)\Big|_{0}^{1}~~~~~~~~~~~~~~~~~~~~~\\\\=1~~~~~~~~~~~~~~~~~&=\frac{1}{2}(e-1)~~~~~~~~~~~~~~~~~~~~~~





\end{array}A​:∫01​xex dx = xex​01​ − ∫01​ex dx      =e − (ex)​01​   =e − (e − 1)=1                 ​B​:  ∫01​xex2 dx =                      Let u=x2. Then, 2du​=xdx= 21​∫01​eu du                 =21​(ex2)​01​                     =21​(e−1)                      ​
Then, 
I=A−B             =1−12(e−1)=32−12eI=A-B \newline{}~~~~~~~~~~~~~=1-\frac{1}{2}(e-1)\newline{}\newline{}
=\frac{3}{2}-\frac{1}{2}eI=A−B             =1−21​(e−1)=23​−21​e

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Evaluate the following double integral:
I=∫04∫x216xsin⁡(y2)dydxI=\int_0^4\int_{x^2}^{16}x\sin(y^2)dydxI=∫04​∫x216​xsin(y2)dydx

I=∫04∫x216xsin⁡(y2)dydx\displaystyle I=\int_0^4\int_{x^2}^{16}x\sin(y^2)dydxI=∫04​∫x216​xsin(y2)dydx 

Change the order of integration to become:
original ordernew order0≤x≤4⇒0≤y≤16x2≤y≤160≤x≤y\begin{array}{ccc}     \textrm{original order}&&\textrm{new order}\\     0\leq x\leq4&\Rightarrow&0\leq y\leq16\\     x^2\leq y\leq 16&&0\leq x\leq \sqrt{y}    \end{array}original order0≤x≤4x2≤y≤16​⇒​new order0≤y≤160≤x≤y​​
Then:
I=∫016∫0yxsin⁡(y2)dxdy     =∫016sin⁡(y2)dy ⋅(x22)∣0y           =∫016sin⁡(y2)dy ⋅(y22 − 0)                                    =12∫016ysin⁡(y2) dy          Let u=y2. Then, du2=ydy=12∫016du2 ⋅ sin⁡u         =14(−cos⁡u)∣y=0y=16            =14(−cos⁡(y2))∣016                     =14(−cos⁡(162) − (−cos⁡02))=14(1−cos⁡(162))          I=\displaystyle\int_0^{16}\int_{0}^{\sqrt{y}}x\sin(y^2)dxdy\newline{}\newline{}
~~~~~
=\int_0^{16}\sin(y^2)dy~\cdot\Bigg(\frac{x^2}{2}\Bigg)\Bigg|_0^{\sqrt{y}}\newline{}\newline{}
~~~~~~~~~~~
=\int_0^{16}\sin(y^2)dy~\cdot\Bigg(\frac{{\sqrt{y}}^2}{2}~-~0\Bigg)\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\frac{1}{2}\int_0^{16}y\sin(y^2)~dy
{\scriptsize\color{magenta}~~~~~~~~~~Let~u=y^2.~Then,~\frac{du}{2}=ydy}\newline{}\newline{}

=\frac{1}{2}\int_0^{16}\frac{du}{2}~\cdot~\sin{u}~~~~~~~~~\newline{}\newline{}

=\frac{1}{4}(-\cos{u})\Big|_{y=0}^{y=16}~~~~~~~~~~~~\newline{}\newline{}

=\frac{1}{4}\Big(-\cos(y^2)\Big)\Big|_0^{16}~~~~~~~~~\newline{}\newline{}

~~~~~~~~~~~~
=\frac{1}{4}\Big(-\cos(16^2)~-~(-\cos0^2)\Big)\newline{}\newline{}

=\frac{1}{4}(1-\cos(16^2))~~~~~~~~~~I=∫016​∫0y​​xsin(y2)dxdy     =∫016​sin(y2)dy ⋅(2x2​)​0y​​           =∫016​sin(y2)dy ⋅(2y​2​ − 0)                                    =21​∫016​ysin(y2) dy          Let u=y2. Then, 2du​=ydy=21​∫016​2du​ ⋅ sinu         =41​(−cosu)​y=0y=16​            =41​(−cos(y2))​016​                     =41​(−cos(162) − (−cos02))=41​(1−cos(162))          

Evaluate ∫D∫(4xy−y3)dA, \displaystyle\int_D\int(4xy-y^3)dA,~∫D​∫(4xy−y3)dA, where D is the region bounded by y=x  and  y=x3.y=\sqrt{x}~~\text{and}~~y=x^3.y=x​  and  y=x3.

Answer

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Evaluate ∫D∫(6x2−40y)dA \displaystyle\int_D\int\Bigg(6x^2-40y\Bigg)dA~∫D​∫(6x2−40y)dA where D is the triangle with vertices (0, 3), (1, 1), and (5, 3). 

Answer

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