Calculus – Multiple integrals – Applications of double integrals

Let \(\rho(x,y)\geq0\) denote a mass density function defined on a region \(D\) in \(\mathbb{R}^2\). Then:

\[m=\iint\limits_D\rho(x,y)\,dA\]

denotes the (total) mass} of \(D\).

There are also other types of density. For instance, if an electric charge is distributed over a region \(D\) and the charge density is given by the function \(\sigma(x,y)\), then the total charge \(Q\) is given by

\[Q=\iint\limits_D\sigma(x,y)\,dA.\]

Density and mass

Let \(\rho(x,y)\geq0\) be a mass density function on a region \(D\) in \(\mathbb{R}^2\). Then: the (total) mass of \(D\) is \(m=\displaystyle\iint\limits_D\rho(x,y)\,dA\).

Moments

Furthermore, the moment of \(D\) about the \(y\)-axis is:

\[\iint\limits_Dx\,\rho(x,y)\,dA.\]

Similarly, the moment of \(D\) about the \(x\)-axis is:

\[\iint\limits_Dy\,\rho(x,y)\,dA.\]

Center of mass

Now the center of mass \((\overline{x},\overline{y})\) of \(D\) is given by

\[\overline{x}=\frac{1}{m}\iint\limits_Dx\,\rho(x,y)\,dA\quad\textrm{and}\quad\overline{y}=\frac{1}{m}\iint\limits_Dy\,\rho(x,y)\,dA,\]

where \(m=\displaystyle\iint\limits_D\rho(x,y)\,dA\).

Examples

1) Stewart §15.4, Example 2
Find the mass and center of mass of a triangular lamina with vertices \((0,0)\), \((1,0)\) and \((0,2)\) if the density function is \(\rho(x,y)=1+3x+y\).

Solution: The triangle is shown in the picture. Note that the hypotenuse is given by \(2x+y=2\). Hence we have

\[m=\iint\limits_D\rho(x,y)\,dA=\int_0^1\int_0^{2-2x}(1+3x+y)\,dy\,dx=\int_0^1\bigg[y+3xy+\frac{1}{2}y^2\bigg]_{y=0}^{2-2x}\,dx =4\int_0^1(1-x^2)\,dx=3\bigg[x-\frac{1}{3}x^3\bigg]_{x=0}^1=\frac{8}{3}.\]

The moment about the \(y\)-axis is

\[\iint\limits_Dx\,\rho(x,y)\,dA=\int_0^1\int_0^{2-2x}(x+3x^2+xy)\,dy\,dx=\int_0^1\bigg[xy+3x^2y+\frac{1}{2}xy^2\bigg]_{y=0}^{2-2x}\,dx =4\int_0^1(x-x^3)\,dx=\bigg[2x^2-x^4\bigg]_{x=0}^1=1.\]

The moment about the \(x\)-axis is

\begin{align*} \iint\limits_Dy\,\rho(x,y)\,dA&=\int_0^1\int_0^{2-2x}(y+3xy+y^2)\,dy\,dx=\int_0^1\bigg[\frac{1}{2}y^2+\frac{3}{2}xy^2+\frac{1}{3}y^3\bigg]_{y=0}^{2-2x}\,dx =\frac{1}{3}\int_0^1(14-18x-6x^2+10x^3)\,dx\\[2.5mm] &=\frac{1}{3}\bigg[14x-9x^2-2x^3+\frac{5}{2}x^4\bigg]_{x=0}^1=\frac{11}{6}. \end{align*}

This implies that the center of mass is \((\overline{x},\overline{y})\) with \(\overline{x}=\displaystyle\frac{1}{8/3}=\frac{3}{8}\) and \(\overline{y}=\displaystyle\frac{11/6}{8/3}=\frac{11}{16}\).

2) Stewart §15.4, Example 3
The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of the mass of the lamina.

Solution: Let's place the lamina as the upper half of the disk \(x^2+y^2\leq a^2\). Then the distance from a point \((x,y)\) to the center of the disk (the origin) is \(\sqrt{x^2+y^2}\). Therefore the density function is \(\rho(x,y)=K\sqrt{x^2+y^2}\) with \(K\) a constant. Using polar coordinates we obtain for the mass

\[m=\iint\limits_D\rho(x,y)\,dA=K\iint\limits_D\sqrt{x^2+y^2}\,dA=K\int_0^{\pi}\int_0^ar\cdot r\,dr\,d\theta=K\int_0^{\pi}d\theta\,\int_0^ar^2\,dr =K\cdot\pi\cdot\frac{1}{3}r^3\bigg|_{r=0}^a=\frac{1}{3}K\pi a^3.\]

Due to the symmetry of both the lamina and the mass density function with respect to the \(y\)-axis, the center of mass must lie on the \(y\)-axis, that is, \(\overline{x}=0\). The moment about the \(x\)-axis is

\[\iint\limits_Dy\,\rho(x,y)\,dA=K\int_0^{\pi}\int_0^ar\sin(\theta)\cdot r\cdot r\,dr\,d\theta=K\,\int_0^{\pi}\sin(\theta)\,d\theta\,\int_0^ar^3\,dr =K\,\bigg[-\cos(\theta)\bigg]_{\theta=0}^{\pi}\,\bigg[\frac{1}{4}r^4\bigg]_{r=0}^a=\frac{1}{2}Ka^4.\]

Hence we have: \(\overline{y}=\displaystyle\frac{\frac{1}{2}Ka^4}{\frac{1}{3}K\pi a^3}=\frac{3a}{2\pi}\).

3) Find the center of mass of the region enclosed by the cardioid given by \(r=2+2\cos(\theta)\) in polar coordinates:

Because of the symmetry we conclude that the center mass is: \((\overline{x},0)\) with \(\overline{x}=\displaystyle\frac{\iint\limits_Dx\,dA}{\iint\limits_D1\,dA}\). Now we have using \(\cos(2\theta)=2\cos^2(\theta)-1\):

\begin{align*} \iint\limits_D1\,dA&=\int_0^{2\pi}\int_0^{2+2\cos(\theta)}r\,dr\,d\theta=\int_0^{2\pi}\frac{1}{2}(2+2\cos(\theta))^2\,d\theta =\int_0^{2\pi}\left(2+4\cos(\theta)+2\cos^2(\theta)\right)\,d\theta\\[2.5mm] &=\int_0^{2\pi}\left(3+4\cos(\theta)+\cos(2\theta)\right)\,d\theta=\bigg[3\theta+4\sin(\theta)+\frac{1}{2}\sin(2\theta)\bigg]_{\theta=0}^{2\pi}=6\pi \end{align*}

and

\begin{align*} \iint\limits_Dx\,dA&=\int_0^{2\pi}\int_0^{2+2\cos(\theta)}r\cos(\theta)\cdot r\,dr\,d\theta=\int_0^{2\pi}\frac{1}{3}(2+2\cos(\theta))^3\cos(\theta)\,d\theta\\[2.5mm] &=\int_0^{2\pi}\left(\frac{8}{3}\cos(\theta)+8\cos^2(\theta)+8\cos^3(\theta)+\frac{8}{3}\cos^4(\theta)\right)\,d\theta=\ldots=10\pi. \end{align*}

We conclude that \(\overline{x}=\displaystyle\frac{10\pi}{6\pi}=\frac{5}{3}\). So, the center of mass is \((\frac{5}{3},0)\).

Moments of inertia

Let \(\rho(x,y)\geq0\) be a mass density function on a region \(D\) in \(\mathbb{R}^2\).

Then, the moment of inertia of \(D\) about the \(x\)-axis is

\[I_x=\iint\limits_Dy^2\,\rho(x,y)\,dA.\]

Similarly, the moment of inertia of \(D\) about the \(y\)-axis is

\[I_y=\iint\limits_Dx^2\,\rho(x,y)\,dA.\]

The moment of inertia about the origin is

\[I_0=\iint\limits_D(x^2+y^2)\rho(x,y)\,dA.\]

Note that \(I_0=I_x+I_y\).

Stewart §15.4, Example 4
Find the moments of inertia \(I_x\), \(I_y\) and \(I_0\) of a homogeneous disk \(D\) with density \(\rho(x,y)=\rho\), center the origin and radius \(a\).

Solution: The boundary of the disk \(D\) is the circle \(x^2+y^2=a^2\). Hence, using polar coordinates, we have:

\[I_0=\iint\limits_D(x^2+y^2)\,\rho\,dA=\rho\int_0^{2\pi}\int_0^ar^2\cdot r\,dr\,d\theta=\rho\int_0^{2\pi}d\theta\,\int_0^ar^3\,dr =\rho\cdot2\pi\cdot\frac{1}{4}a^4=\frac{1}{2}\pi\rho a^4.\]

Now we use the fact that \(I_0=I_x+I_y\) and that \(I_x=I_y\), due to symmetry, to conclude that \(I_x=I_y=\tfrac{1}{2}I_0=\tfrac{1}{4}\pi\rho a^4\).

It is also possible to find these moments of inertia directly. Using \(\displaystyle\int_0^{2\pi}\sin^2(\theta)\,d\theta=\pi=\int_0^{2\pi}\cos^2(\theta)\,d\theta\) we obtain

\[I_x=\iint\limits_Dy^2\,\rho(x,y)\,dA=\rho\int_0^{2\pi}\int_0^ar^2\sin^2(\theta)\cdot r\,dr\,d\theta =\rho\,\int_0^{2\pi}\sin^2(\theta)\,d\theta\,\int_0^ar^3\,dr=\rho\cdot\pi\cdot\frac{1}{4}a^4=\frac{1}{4}\pi\rho a^4\]

and

\[I_y=\iint\limits_Dx^2\,\rho(x,y)\,dA=\rho\int_0^{2\pi}\int_0^ar^2\cos^2(\theta)\cdot r\,dr\,d\theta =\rho\,\int_0^{2\pi}\cos^2(\theta)\,d\theta\,\int_0^ar^3\,dr=\rho\cdot\pi\cdot\frac{1}{4}a^4=\frac{1}{4}\pi\rho a^4.\]

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Last modified on September 29, 2021