Wize University Calculus 3 Textbook > Multiple Integrals

Double Integrals over Rectangular Regions

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Double Integrals over Rectangular Regions
The Definite Integral
The definite integral for a function y=f(x)y=f(x)y=f(x) represents the area under a curve bounded by the region f(x)f(x)f(x), the y-axis, and x=a, x=bx=a,~x=bx=a, x=b, where a,b∈Ra,b\in\mathbb{R}a,b∈R.
The definite integral for a function z = f(x, y) represents the volume of a solid that lies below f(x, y) and above a region D in the XY-plane, where D is defined as D={(x,y)∣a≤x≤b, c≤y≤d}D = \{(x, y)|a \le x \le b,\ c \le y \le d\}D={(x,y)∣a≤x≤b, c≤y≤d}

Therefore, we can say that the volume under a function f(x, y) over a rectangular area xij∗yij∗x_{ij}^*y_{ij}^*xij∗​yij∗​is f(xij∗,yij∗)ΔA\color{blue}f(x_{ij}^*,y_{ij}^*)\Delta Af(xij∗​,yij∗​)ΔAand can be expressed as:
V=∬Df(x,y)dA=lim⁡m,n→∞∑i=1n∑j=1mf(xij∗,yij∗)ΔA\color{blue}V=\iint_Df(x,y)dA=\lim_{m,n\to\infty}\sum_{i=1}^n\sum_{j=1}^mf(x_{ij}^*,y_{ij}^*)\Delta AV=∬D​f(x,y)dA=m,n→∞lim​i=1∑n​j=1∑m​f(xij∗​,yij∗​)ΔA
where f(x, y) is integrable over D if the limit exists.
Fubini's Theorem: If f(x,y)f(x,y)f(x,y) is continuous over the rectangular region R={(x,y)∣a≤x≤b, c≤y≤d}R=\{(x,y)|a\le x\le b,\ c\le y\le d\}R={(x,y)∣a≤x≤b, c≤y≤d}, then:
∬Rf(x,y)dA=∫cd∫abf(x,y)dxdy=∫ab∫cdf(x,y)dydx\color{blue}\iint_Rf(x,y)dA=\int_c^d\int_a^bf(x,y)dxdy=\int_a^b\int_c^df(x,y)dydx∬R​f(x,y)dA=∫cd​∫ab​f(x,y)dxdy=∫ab​∫cd​f(x,y)dydx

Wize Concept
To perform double integrals, we follow a similar procedure we had for partial derivatives:
    i) To solve double integral with respect to xxx, all terms containing yyy is treated to be constant.
    ii) To solve double integral with respect to yyy, all terms containing xxx is treated to be constant.

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Evaluate ∫01∫12x2y3dydx\displaystyle\int_{0}^1\int_1^2x^2y^3dydx∫01​∫12​x2y3dydx

∫01(∫12x2y3 dy)dx=∫01x2 dx (y44)∣12                                            =∫01x2 dx(244−14)                                 =154×(x33)∣01             =54\displaystyle\int_0^1\Bigg(\int_1^2x^2y^3~dy\Bigg)dx=\int_0^1x^2~dx~\Big(\dfrac{y^4}{4}\Big)\bigg\vert_1^2\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle=\int_0^1x^2~dx\Bigg(\frac{2^4}{4}-\dfrac{1}{4}\Bigg)\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\dfrac{15}{4}\times\Bigg(\dfrac{x^3}{3}\Bigg)\bigg\vert_0^1\newline{}\newline{}
~~~~~~~~~~~~~=\dfrac{5}{4}∫01​(∫12​x2y3 dy)dx=∫01​x2 dx (4y4​)​12​                                            =∫01​x2 dx(424​−41​)                                 =415​×(3x3​)​01​             =45​

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Compute ∫03∫02xy2x2+y3 dydx\displaystyle\int_0^3\int_0^2xy^2\sqrt{x^2+y^3}~dydx∫03​∫02​xy2x2+y3​ dydx

First, integrate with respect to y using substituion. 

Let u=x2+y3  →  du=3y2dyu=x^2+y^3~~{\color{red}\rightarrow}~~du=3y^2dy
u=x2+y3  →  du=3y2dy. Then:
∫03∫02x2+y3xy2 dydx = ∫03x dx∫02x2+y3 y2dy                                          =∫03x dx∫02u ⋅13 ⋅ du                                                                                                   =∫03x dx(29u32)∣y=0y=2       ∗∗∗Do not forget to substitute y back in for u                                            =29∫03x dx((x2+y3)32)∣02                                              =29∫03x dx((x2+8)32−x3)                                                        =29(∫03x(x2+8)32 − ∫03x4 dx)\displaystyle\int_0^3\int_0^2\sqrt{x^2+y^3}xy^2~dydx~=~\int_0^3 x~dx\int_0^2\sqrt{x^2+y^3}~y^2dy\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\int_0^3 x~dx\int_0^2\sqrt{u}~\cdot\frac{1}{3}~\cdot~du
\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\int_0^3x~dx\Bigg(\frac{2}{9}u^{\frac{3}{2}}\Bigg)\Bigg|_{y=0}^{y=2}~~~~~~~{\scriptsize{\color{magenta}***Do ~not~forget~to~substitute~y~back~in~for~u}}\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\frac{2}{9}\int_0^3x~dx\Bigg((x^2+y^3)^{\frac{3}{2}}\Bigg)\Bigg|_0^2\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\frac{2}{9}\int_0^3x~dx\Big((x^2+8)^{\frac{3}{2}}-x^3\Big)\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\frac{2}{9}\Bigg(\int_0^3x(x^2+8)^{\frac{3}{2}}~-~\int_0^3x^4~dx\Bigg)∫03​∫02​x2+y3​xy2 dydx = ∫03​x dx∫02​x2+y3​ y2dy                                          =∫03​x dx∫02​u​ ⋅31​ ⋅ du                                                                                                   =∫03​x dx(92​u23​)​y=0y=2​       ∗∗∗Do not forget to substitute y back in for u                                            =92​∫03​x dx((x2+y3)23​)​02​                                              =92​∫03​x dx((x2+8)23​−x3)                                                        =92​(∫03​x(x2+8)23​ − ∫03​x4 dx)
Let u=x2+8  →   du2=xdxu=x^2+8~~{\color{red}\rightarrow}~~~\frac{du}{2}=xdxu=x2+8  →   2du​=xdx. Then:
29(∫03u322 du − ∫03x4 dx)= 29(u525∣03 − x55∣03)                                                            =245((x2+8)52∣03 −x5∣03)                                          =245(1752−35)                               ≈34.1137\frac{2}{9}\Bigg(\int_0^3\frac{u^{\frac{3}{2}}}{2}~du~-~\int_0^3x^4~dx\Bigg)

=~\frac{2}{9}\Bigg(\frac{u^{\frac{5}{2}}}{5}\Bigg|_0^3~-~\frac{x^5}{5}\Bigg|_0^3\Bigg)\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\frac{2}{45}\Bigg((x^2+8)^{\frac{5}{2}}\Bigg|_0^3~-x^5\Bigg|_0^3
\Bigg)\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
=\frac{2}{45}\Big(17^{\frac{5}{2}}-3^5\Big)\newline{}\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\approx34.113792​(∫03​2u23​​ du − ∫03​x4 dx)= 92​(5u25​​​03​ − 5x5​​03​)                                                            =452​((x2+8)25​​03​ −x5​03​)                                          =452​(1725​−35)                               ≈34.1137

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Determine the volume that lies under f(x, y)=xycos(x2y)dAf(x,~y)=xycos(x^2y)dAf(x, y)=xycos(x2y)dA and above the rectangle R: [-2, 3] x [-1, 1].

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