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Wize University Calculus 3 Textbook > Multiple Integrals

Double Integrals over Rectangular Regions

Introduction to Multiple IntegralsPrevious Section
Double Integrals over General RegionsNext Section
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Double Integrals over Rectangular Regions

The Definite Integral

  • The definite integral for a function y=f(x)y=f(x) represents the area under a curve bounded by the region f(x)f(x), the y-axis, and x=a, x=bx=a,~x=b, where a,b∈Ra,b\in\mathbb{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\}

Therefore, we can say that the volume under a function f(x, y) over a rectangular area xij∗yij∗x_{ij}^*y_{ij}^*is f(xij∗,yij∗)ΔA\color{blue}f(x_{ij}^*,y_{ij}^*)\Delta Aand 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 A
where f(x, y) is integrable over D if the limit exists.
Fubini's Theorem: If 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\}, 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

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

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

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Compute ∫03∫02xy2x2+y3 dydx\displaystyle\int_0^3\int_0^2xy^2\sqrt{x^2+y^3}~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 . 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)
Let u=x2+8  →   du2=xdxu=x^2+8~~{\color{red}\rightarrow}~~~\frac{du}{2}=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{} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \approx34.1137

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