Wize University Calculus 3 Textbook > Multiple Integrals

Application of the Double Integral: Mass, Center of Mass, Surface Area

Double Integrals in Polar CoordinatesPrevious Section

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Double integrals can be used to calculate many physical quantities including mass, moments respect to an axis, centre of mass coordinates, moments of inertia, and surface area. 

Mass of a plate: If the surface mass density (mass per unit area) of a plate over the area DDD is defined as ρ(x,y)\rho(x,y)ρ(x,y), the total mass of the plate is calculated as follows:
m=∬Dρ(x,y)dAm=\iint_D\rho(x,y)dAm=∬D​ρ(x,y)dA
Other properties of the plate can be determined as follows:
i) Moment of a plate with respect to either xxx or yyy axis: If we denote MxM_xMx​ and MyM_yMy​ the moment of the plate about xxx and yyy axis, we will have:
Mx=∬Dyρ(x,y)dAM_x=\iint_Dy\rho(x,y)dAMx​=∬D​yρ(x,y)dA

here yyy is the distance of the point of interest to the xxx-axis
My=∬Dxρ(x,y)dAM_y=\iint_Dx\rho(x,y)dAMy​=∬D​xρ(x,y)dA

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here xxx is the distance of the point of interest to the yyy-axis
ii) The coordinates (xˉ,yˉ)(\bar x,\bar y)(xˉ,yˉ​) of the center of mass: The center of mass is the point at which all of the mass of the object is concentrated.
If the force is applied to the center of mass, there is no torque generated in the object.
The coordinates of this point could be calculated as follows:
xˉ=Mym=1m∬Dxρ(x,y)dAyˉ=Mxm=1m∬Dyρ(x,y)dA\begin{array}{l}
\displaystyle\bar x=\dfrac{M_y}{m}=\dfrac{1}{m}\iint_Dx\rho(x,y)dA\\[10pt]
\displaystyle\bar y=\dfrac{M_x}{m}=\dfrac{1}{m}\iint_Dy\rho(x,y)dA
\end{array}xˉ=mMy​​=m1​∬D​xρ(x,y)dAyˉ​=mMx​​=m1​∬D​yρ(x,y)dA​
where the mass mmm is given by:
m=∬Dρ(x,y)dAm=\iint_D\rho(x,y)dAm=∬D​ρ(x,y)dA

iii) The moment of inertia of plate about xxx axis (IxI_xIx​) and about yyy axis (IyI_yIy​): Moment of inertia is the rotational analog to the mass!
Ix=∬Dy2ρ(x,y)dAI_x=\iint_Dy^2\rho(x,y)dAIx​=∬D​y2ρ(x,y)dA

here yyy is the distance of the point of interest to the xxx-axis
Iy=∬Dx2ρ(x,y)dAI_y=\iint_Dx^2\rho(x,y)dAIy​=∬D​x2ρ(x,y)dA

here xxx is the distance of the point of interest to the yyy-axis
iv) The polar moment of inertia (I0I_0I0​): The polar moment of inertia is calculated as follows:
I0=∬D(x2+y2)ρ(x,y)dA=Ix+IyI_0=\iint_D(x^2+y^2)\rho(x,y)dA=I_x+I_yI0​=∬D​(x2+y2)ρ(x,y)dA=Ix​+Iy​

The surface area of a surface z = f(x, y) where (x, y) is a point from the region D in the xy - plane, can be defined as:
SA=∫∫fx2+fy2+1 dAD                                 SA=\displaystyle\int\int\sqrt{f_x^2+f_y^2+1}~dA\newline{}
{\scriptsize{D}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~SA=∫∫fx2​+fy2​+1​ dAD                                 

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Given the triangle with vertices (0, 0), (0, 5), (5, 0) and a density defined by ρ(x, y)=yx\rho(x,~y)=y\sqrt{x}ρ(x, y)=yx​, find the mass. 

Find m:
m=∫05∫05−xyx dydx=∫05x dx∫05−xy dy=∫05x dx(y22)∣05−x=12∫05(x52−10x32+25x12)dx=12(2x727−4x52+50x323)∣05=x727−2x52+25x323∣05=200521m=\displaystyle\int_0^5\int_0^{5-x}y\sqrt{x}~dydx\newline{}\newline{}

=\int_0^5\sqrt{x}~dx\int_0^{5-x}y~dy\newline{}\newline{}

=\int_0^5\sqrt{x}~dx\Bigg(\frac{y^2}{2}\Bigg)\Bigg|_0^{5-x}\newline{}\newline{}

=\frac{1}{2}\int_0^5\Big(x^{\frac{5}{2}}-10x^{\frac{3}{2}}+25x^{\frac{1}{2}}\Big)dx\newline{}\newline{}

=\frac{1}{2}\Bigg(\frac{2x^\frac{7}{2}}{7}-4x^\frac{5}{2}+\frac{50x^\frac{3}{2}}{3}\Bigg)\Bigg|_0^5\newline{}\newline{}

=\frac{x^\frac{7}{2}}{7}-2x^\frac{5}{2}+\frac{25x^\frac{3}{2}}{3}\Bigg|_0^5\newline{}\newline{}

=\frac{200\sqrt{5}}{21}m=∫05​∫05−x​yx​ dydx=∫05​x​ dx∫05−x​y dy=∫05​x​ dx(2y2​)​05−x​=21​∫05​(x25​−10x23​+25x21​)dx=21​(72x27​​−4x25​+350x23​​)​05​=7x27​​−2x25​+325x23​​​05​=212005​​

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Find the center of mass of a semicircular plate with a constant density of BBB and a radius of aaa.

We use polar coordinates system!


m=∬Dρ(x,y)dA=∬DBdA=B∬DdA\displaystyle m=\iint_D \rho(x,y)dA=\iint_D BdA=B\iint_D dAm=∬D​ρ(x,y)dA=∬D​BdA=B∬D​dA
m=B∫0πdθ∫0a(1)rdr=B∫0πdθ⋅12r2∣0a\displaystyle m=B\int_0^{\pi} d\theta\int_0^a (1)rdr=B\int_0^{\pi}d\theta\cdot \frac{1}{2}r^2\bigg\vert_0^am=B∫0π​dθ∫0a​(1)rdr=B∫0π​dθ⋅21​r2​0a​
=B∫0πdθ12a2=Ba22∫0πdθ=Ba2π2\displaystyle=B\int_0^\pi d\theta\frac{1}{2}a^2=\frac{Ba^2}{2}\int_0^\pi d\theta=\frac{Ba^2\pi}{2}=B∫0π​dθ21​a2=2Ba2​∫0π​dθ=2Ba2π​
xˉ=Mym=1m∬xρ(x,y)  dA\displaystyle\bar{x}=\frac{M_y}{m}=\frac{1}{m}\iint x\rho(x,y)\;dAxˉ=mMy​​=m1​∬xρ(x,y)dA

My=∬xρ(x,y)  dA=∫0πdθ∫0arcos⁡(θ)Br  dr\displaystyle M_y=\iint x\rho(x,y)\;dA=\int_0^{\pi} d\theta\int_0^a r\cos(\theta)Br\;drMy​=∬xρ(x,y)dA=∫0π​dθ∫0a​rcos(θ)Brdr=B∫0πdθ∫0ar2cos⁡(θ)  dr\displaystyle=   B\int_0^{\pi} d\theta\int_0^a r^2\cos(\theta)\;dr=B∫0π​dθ∫0a​r2cos(θ)dr
=B∫0πdθcos⁡(θ)13r3∣0a=Ba33∫0πcos⁡(θ)  dθ=Ba33sin⁡(θ)∣0π\displaystyle=B\int_0^\pi d\theta\cos(\theta)\frac{1}{3}r^3\bigg\vert_0^a=    \frac{Ba^3}{3}\int_0^\pi\cos(\theta)\; d\theta=\frac{Ba^3}{3}\sin(\theta)\bigg\vert_0^\pi=B∫0π​dθcos(θ)31​r3​0a​=3Ba3​∫0π​cos(θ)dθ=3Ba3​sin(θ)​0π​
=Ba33[sin⁡(pi)−sin⁡(0)]=0⇒xˉ=Mym=0\displaystyle=\frac{Ba^3}{3}[\sin(pi)-\sin(0)]=0\Rightarrow \bar{x}=\frac{M_y}{m}=0=3Ba3​[sin(pi)−sin(0)]=0⇒xˉ=mMy​​=0

Mx=∬Dyρ(x,y)  dA=∫0πdθ∫0arsin⁡(θ)Brdr=B∫0πdθsin⁡(θ)13r3∣0a\displaystyle M_x=\iint_D y\rho(x,y)\;dA=\int_0^{\pi} d\theta\int_0^a r\sin(\theta)Brdr= B\int_0^\pi d\theta\sin(\theta)\frac{1}{3}r^3\bigg\vert_0^aMx​=∬D​yρ(x,y)dA=∫0π​dθ∫0a​rsin(θ)Brdr=B∫0π​dθsin(θ)31​r3​0a​
=Ba33∫0πsin⁡(θ)  dθ=Ba33(−cos⁡(θ))∣0π\displaystyle=\frac{Ba^3}{3}\int_0^\pi\sin(\theta)\; d\theta=\frac{Ba^3}{3}(-\cos(\theta))\bigg\vert_0^\pi=3Ba3​∫0π​sin(θ)dθ=3Ba3​(−cos(θ))​0π​
=Ba33[−cos⁡(π)−(−cos⁡0)]\displaystyle=\frac{Ba^3}{3}[-\cos(\pi)-{(-\cos0)}]=3Ba3​[−cos(π)−(−cos0)]=2Ba33⇒yˉ=Mxm=2Ba33Ba2π2=43aπ\displaystyle=\frac{2Ba^3}{3}\Rightarrow \bar{y}=\frac{M_x}{m}=\frac{\frac{2Ba^3}{3}}{\frac{Ba^2\pi}{2}}=\frac{4}{3}\frac{a}{\pi}=32Ba3​⇒yˉ​=mMx​​=2Ba2π​32Ba3​​=34​πa​

Find the surface area of the part of the plane x+y+z=3x+y+z=3x+y+z=3 that is in the first octant. 

Letting z=0z=0z=0 shows us that the bounds for x and y in the first octant can be determined by looking at x+y=3x+y=3x+y=3.  Let the domain, D, be defined as: D:{(x, y)∣ 0≤x≤3; 0≤y≤3−x, z=3−x−y}D:\{(x,~y)|~0\leq{x}\leq{3};~0\leq{y}\leq{3-x},~z=3-x-y\}D:{(x, y)∣ 0≤x≤3; 0≤y≤3−x, z=3−x−y}. Then, since f(x,y)=3−x−yf(x,y)=3-x-yf(x,y)=3−x−y we have the below partial derivatives and surface area integral:
fx=−1fy=−1SA=∫03∫03−x(−1)2+(−1)2+1 dydx=3∫03(3−x) dx=3(3x−12x2)∣03=3(9−92)=932f_x=-1\newline{}
f_y=-1\newline{}\newline{}
SA=\int_0^3\int_0^{3-x}\sqrt{(-1)^2+(-1)^2+1}~dydx\newline{}\newline{}
=\sqrt{3}\int_0^3\Bigg(3-x\Bigg)~dx\newline{}\newline{}
=\sqrt{3}\Bigg(3x-\frac12x^2\Bigg)\Bigg|_0^3\newline{}\newline{}
=\sqrt{3}(9-\frac92)\newline{}\newline{}
=\frac{9\sqrt{3}}2fx​=−1fy​=−1SA=∫03​∫03−x​(−1)2+(−1)2+1​ dydx=3​∫03​(3−x) dx=3​(3x−21​x2)​03​=3​(9−29​)=293​​

Find the center of mass between y = cosx and the x-axis and between x = −π2-\frac{\pi}{2}−2π​ and x = π2\frac{\pi}{2}2π​, where ρ(x,y)=1.\rho(x,y)=1.ρ(x,y)=1.

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Find the center of mass of a 2D plate in the first quadrant for the circle x2+y2≤1x^2+y^2\leq1x2+y2≤1 and ρ(x, y)=A(x2+y2).\rho(x,~y)=A(x^2+y^2).ρ(x, y)=A(x2+y2).

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Determine the surface area of the portion y=x2+z2−1y=x^2+z^2-1y=x2+z2−1 inside x2+z2=4.x^2+z^2=4.x2+z2=4.

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Wize University Calculus 3 Textbook > Multiple Integrals

Application of the Double Integral: Mass, Center of Mass, Surface Area

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