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Chapter 15

15.1

Double Integrals

In 2d

(15.1#1)

Area = f(xi)Δx\sum f(x_{i})\cdot \Delta x

in 3d

xy-plane is 'bounding plane'

Gets square - Pick corner or centre

V=f(xi,yj)ΔxΔy=i=1nj=1mf(xi,yj)ΔxΔy \begin{align} V &= \sum f(x_{i},y_{j})\Delta x\Delta y \\ \\ &= \sum_{i=1}^n \sum_{j=1}^m f(x_{i},y_{j})\Delta x\Delta y \end{align}

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Ex (15.1#3)

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R=[0,2]xx[0,2]y R=\underbrace{ [0,2] }_{ x }x\underbrace{ [0,2] }_{ y }

Ex (15.1#4)

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R={(x,y)1x1,2y2}=[1,1]x[2,2] R=\{(x,y)|-1\leq x\leq{1},-2\leq y\leq{2}\}=[-1,1]x[-2,2]

Double Integral

To Eval Rf(x,y)dAdxdy \begin{align} \int \int _R f(x,y) \underbrace{ dA }_{ dxdy } \end{align}

first eval the "inner" in _ with respect to the inner diff holding the outer diff var as a constant then eval the outer integral

The outer limits of integration ==MUST BE CONSTANTS==

cdabf(x,y)dxdy \begin{align} \int _c ^d \int _{a} ^b f(x,y) \, dx \, dy \\ \end{align}

(15.1#5) 0202(16x22y2)dydx=48 \begin{align} \int _{0} ^2 \int _{0} ^2 (16-x^{2}-2y^{2}) \, dy \, dx \\ =48 \end{align}

15.2

Ex 1

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(15.2#1)

B/c vertical

xBxTyByTf(x)dydxyT=1+x2yB=2x2for a and b2x2=1+x2x2=1x=±1112x21+x2(x+2y)dydx[35x514x4+23x3+12x2+1]11=65+43+2 \begin{align} \int _{x_{B}} ^{x_{T}} \int _{y_{B}} ^{y_{T}} f(x) \, dy \, dx \\ y_{T} &= 1 + x^{2} \\ y_{B} &= 2x^{2} \\ \\ \text{for } a \text{ and } b \\ 2x^{2} &= 1+x^{2} \\ x^{2} &= 1 \\ x &= \pm 1 \\ \\ \int _{-1} ^1 \int _{2x^{2}} ^{1+x^{2}} (x+2y )\, dy \, dx \\ \left[ -\frac{3}{5}x^5 - \frac{1}{4}x^4 + \frac{2}{3} x ^3 + \frac{1}{2}x^{2} + 1 \right] _{-1} ^1 = -\frac{6}{5}+\frac{4}{3}+2 \end{align}

Ex 2

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(15.2#2)

Ex 3

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(15.2#3)

15.3

If I see circles

x=rcosθy=rsinθx2+y2=r20ra0θ2πSlength=rdθdA=dr(rdθ)=rdrdθ \begin{align} x&=r\cos \theta \\ y&=r\sin \theta \\ x^{2}+y^{2}&=r^{2} \\ \\ \\ 0 &\leq r \leq a \\ 0 &\leq \theta \leq 2\pi \\ \\ S_{\text{length}} &= r d \theta \\ dA &= dr(rd\theta) = r dr d\theta \end{align}

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x=rcosθy=rsinθx2+y2=r2dA=rdrdθ1r20θπ0π12[3rcosθ+4r2sin2θ]rdrdθ0π12[3r2cosθ+4r3sin2θ]drdθ[7sinθ]0π+152[1cos2θ]dθ \begin{align} x &= r\cos \theta \\ y &= r\sin \theta \\ x^{2}+y^{2}&=r^{2} \\ dA &= r dr d\theta \\ \\ \\ 1 &\leq r \leq 2 \\ 0 &\leq \theta \leq \pi \\ \\ \\ \int _{0} ^\pi \int _{1} ^2 &\left[3r\cos \theta+4r^{2}\sin ^{2}\theta\right]r \, dr \, d\theta \\ \\ \int _{0} ^\pi \int _{1} ^2 &\left[3r^{2}\cos \theta+4r^{3}\sin ^{2}\theta\right] dr \, d\theta \\ \\ \\ \\ [7\sin \theta]_0 ^\pi &+ \int \frac{15}{2}[1-\cos2\theta] \, d\theta \end{align}

Important knowledge: sin2θ1cos(2θ)2 \sin ^{2}\theta \to \frac{1-\cos(2\theta)}{2}

Ex

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Circle on xyxy-plane

1x2y2dA02π01[1r2]rdrdθ02π01[rr3]drdθ02π[1214]drdθ2π[14]=π2 \begin{align} \iint 1-x^{2}-y^{2} \, dA \\ \int _{0} ^{2\pi} \int _{0} ^1\left[ 1-r^{2} \right]r \, dr \, d\theta \\ \int _{0} ^{2\pi} \int _{0} ^1\left[ r-r^{3} \right] \, dr \, d\theta \\ \int _{0} ^{2\pi}\left[ \frac{1}{2}-\frac{1}{4} \right] \, dr \, d\theta \\ 2\pi\left[ \frac{1}{4} \right] = \frac{\pi}{2} \end{align}

Ex 2

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Bounding Region: ![[Pasted image 20221003083916.png|100]]

0r90θ2π02π09(r2)72dxdx02π09r8dxdx=2π(98) \begin{align} 0\leq r\leq 9 \\ 0\leq \theta \leq 2\pi \\ \\ \int _{0} ^{2\pi} \int _{0} ^9 (r^{2})^{\frac{7}{2}} \, dx \, dx \\ \int _{0} ^{2\pi} \int _{0} ^9 r^8 \, dx \, dx \\ = 2\pi(9^8) \end{align}

Ex 3

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A=RdAA=Drdrdθr218rcosθ=0r(r18cosθ)=0r=0r=18cosθr1r1802π18cosθ18drdθ=81π2 \begin{align} \\ A =\iint _{R} dA \\ A =\iint _{D} r \, dr \, d\theta \\ \\ \\ \\ \\ r^{2}-18r\cos \theta&=0 \\ r(r-18\cos \theta)&=0 \\ r&=0 & r=18\cos \theta \\ r_{1} &\leq r \leq 18 \\ \\ \\ \int _{0} ^{2\pi} \int _{18\cos \theta} ^{18} \, dr \, d\theta \\ = \frac{81\pi}{2} \end{align}

Ex 4

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v=dAdzz2=100717(x2+y2)z=±100717(x2+y2)v=[2100717(x2+y2)]dA=202π010[100717r2]rdrdθ \begin{align} v&=\int \int \int dA \, dz \\ z^{2}&=\frac{100}{7}-\frac{1}{7}(x^{2}+y^{2}) \\ z&=\pm\sqrt{\frac{100}{7}-\frac{1}{7}(x^{2}+y^{2})} \\ \\ \\ \\ v&=\int \int \left[ 2\sqrt{\frac{100}{7}-\frac{1}{7}(x^{2}+y^{2})} \right] \, dA \\ \\ &= 2\int _{0} ^{2\pi} \int _{0} ^{10} \left[ \sqrt{ \frac{100}{7}-\frac{1}{7}r^{2} } \right]r \, dr \, d\theta \end{align}

15.4

density: ρ\rho = md\frac{m}{d} -> m=ρA=Rρ(x,y)dA m=\rho A= \int \int _{R} \rho(x,y) \, dA

Ex 1

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m=Rρ(x,y)dA=01022x[1+3x+y]dydx=01[1y3xy12y2]022xdx=01[1(22x)3x(22x)12(22x)2]dx=83 \begin{align} m &= \int \int _{R} \rho(x,y) \, dA \\ &=\int _{0} ^ 1 \int _{0} ^{2-2x} [1+3x+y] \, dydx \\ &= \int _{0} ^ 1 \left[ 1y-3xy-\frac{1}{2}y^2 \right]_{0} ^{2-2x} \, dx \\ &= \int _{0} ^1 \left[ 1(2-2x) -3x(2-2x) -\frac{1}{2}(2-2x)^2 \right] \, dx \\ &= \frac{8}{3} \end{align}

Moments

Mx=Ryρ(x,y)dAMy=Rxρ(x,y)dACentre of mass=(x,y)x=Mymy=Mxm \begin{align} \\ M_{x} &= \int \int _{R} y \rho(x,y) \, dA \\ M_{y} &= \int \int _{R} x\rho(x,y) \, dA \\ \text{Centre of mass} &= (\overline{x},\overline{y}) \\ \overline{x}&=\frac{M_{y}}{m} \\ \overline{y}&=\frac{M_{x}}{m} \end{align}

Using Example 1 from above: m=83ρ(x,y)=1+3x+yx=1mRxρ(x,y)dA=38y==1116 \begin{align} m&=\frac{8}{3} \\ \rho(x,y) &=1+3x+y \\ \\ \\ \overline{x}&=\frac{1}{m}\int \int _{R} x\rho(x,y) \, dA \\ &= \frac{3}{8} \\ \\ \\ \overline{y} &= \dots \\ &= \frac{11}{16} \end{align}

Ex 2

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2nd Moment (Inertia)

Ix=Dy2ρ(x,y)dAIy=Dx2ρ(x,y)dAIo=D(x2+y2)ρ(x,y)dA \begin{align} I_{x} &= \int \int _{D} y^2 \rho(x,y) \, dA \\ I_{y} &= \int \int _{D} x^2 \rho(x,y) \, dA \\ I_{o} &= \int \int _{D} (x^2 + y^2)\rho(x,y) \, dA \\ \\ \end{align}

Ex

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15.6

Triple Integrals in Cartesian coords

Ef(x,y,z)dVvolume=EdV \begin{align} \\ &\int \int \int _{E} f(x,y,z) \, dV \\ \text{volume} &= \int \int \int _{E} \, dV \end{align}

Ex

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=03xyz2dzdA=xy[z33]03dA=9xydA=90112xydydx==274 \begin{align} &=\int \int \int _{0} ^3 xyz^2 \, dz \, dA \\ &= \int \int xy\left[ \frac{z^3}{3} \right]_{0} ^3 \, dA \\ &= 9\int \int xy \, dA \\ &= 9\int _{0} ^ 1\int _{-1} ^2 xy \, dy \, dx \\ &=\dots \\ &= \frac{27}{4} \end{align}

Ex 2

EzdVE:x=0,y=0,z=0,x+y+z=1=01xyzdzdA=12R(1xy)2dA=120101x(1xy)2dydx==124 \begin{align} &\int \int \int _{E}z \, dV \\ E&: x=0,\quad y=0,\quad z=0,\quad x+y+z=1 \\ &=\int \int \int _{0} ^{1-x-y} z \, dz \, dA \\ &=\frac{1}{2} \int \int _{R} (1-x-y)^2 \, dA \\ &= \frac{1}{2}\int _{0} ^1 \int _{0} ^{1-x}(1-x-y)^2 \, dy \, dx \\ &= \dots \\ &= \frac{1}{24} \end{align}

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Ex 3

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x2+y24x2+y2dzdA==Rx2+y2(4(x2+y2))dA=02π02r(4r2)rdrdθ \begin{align} &\int \int \int _{x^2+y^2}^4\sqrt{ x^2+y^2 } \, dz \, dA \\ &= \dots \\ &= \int \int _{R} \sqrt{ x^2+y^2 } (4-(x^2+y^2)) \, dA \\ &= \int _{0} ^{2\pi} \int _{0} ^2 r(4-r^2)r \, dr \, d\theta \end{align}

Ex 4

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15.7 - Cylindrical Coordinates

Do I see circles in only one plane or do I see cylinders

let z=zx=rcosθy=rsinθdv=rdzdrdθx2+y2=r2 \begin{align} \text{let } \\ z&=z \\ x&= r\cos \theta \\ y&=r\sin \theta \\ dv&=r\,dz\,dr\,d\theta \\ x^{2}+y^{2}&=r^{2} \end{align}

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Ex 1

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Bounding Region

x2+y2z24x2y4x22x222(x2+y2)(2x2+y2)dxdx \begin{align} \sqrt{ x^{2}+y^{2} } &\leq z\leq 2 \\ -\sqrt{ 4-x^{2} } &\leq y \leq \sqrt{ 4-x^{2} } \\ -2 &\leq x \leq 2 \\ \\ \int _{-2} ^2 \int _{-\sqrt{ \quad }} ^{\sqrt{ \quad }} (x^{2}+y^{2})(2-\sqrt{ x^{2}+y^{2} }) \, dx \, dx \end{align}

Ex 2

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v=dvv=4(x2+y2)4(x2+y2)dzdA \begin{align} v&=\int \int \int \, dv \\ v&=\int \int \int _{-\sqrt{ 4-(x^{2}+y^{2}) }} ^{\sqrt{ 4-(x^{2}+y^{2}) }} dz\,dA \\ \\ \end{align}

Ex 3

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v=R01/(x2+y2)14dzdA=R(x2+y2)14dA=02π1r27drdθ=2π[126r26]1=2π26[01]=π13 \begin{align} v&=\int \int _R \int ^{ 1/(x^{2}+y^{2})^{14} }_{0}dzdA \\ &= \int \int _{R} (x^{2}+y^{2})^{-14} \, dA \\ &= \int _{0}^{2\pi}\int _{1}^\infty r^{-27} \, dr \,d\theta \\ &= 2\pi\left[ \frac{1}{-26}r^{-26} \right]_{1}^\infty \\ &= -\frac{2\pi}{26}[0-1] \\ &= \frac{\pi}{13} \end{align}