Select Page

Double Integrals

# Physical Applications of Double Integrals

Page 1
Problem 1
Page 2
Problems 2-3

### Mass and Static Moments of a Lamina

Suppose we have a lamina which occupies a region $$R$$ in the $$xy$$-plane and is made of non-homogeneous material. Its density at a point $$\left( {x,y} \right)$$ in the region $$R$$ is $$\rho \left( {x,y} \right).$$ The total mass of the lamina is expressed through the double integral as follows:
$m = \iint\limits_R {\rho \left( {x,y} \right)dA} .$ The static moment of the lamina about the $$x$$-axis is given by the formula
${M_x} = \iint\limits_R {y\rho \left( {x,y} \right)dA} .$ Similarly, the static moment of the lamina about the $$y$$-axis is
${M_y} = \iint\limits_R {x\rho \left( {x,y} \right)dA} .$ The coordinates of the center of mass of a lamina occupying the region $$R$$ in the $$xy$$-plane with density function $$\rho \left( {x,y} \right)$$ are described by the formulas
${\bar x = \frac{{{M_y}}}{m} } = {\frac{1}{m}\iint\limits_R {x\rho \left( {x,y} \right)dA} } = {\frac{{\iint\limits_R {x\rho \left( {x,y} \right)dA} }}{{\iint\limits_R {\rho \left( {x,y} \right)dA} }},}$ ${\bar y = \frac{{{M_x}}}{m} } = {\frac{1}{m}\iint\limits_R {y\rho \left( {x,y} \right)dA} } = {\frac{{\iint\limits_R {y\rho \left( {x,y} \right)dA} }}{{\iint\limits_R {\rho \left( {x,y} \right)dA} }}.}$ When the mass density of the lamina is $$\rho \left( {x,y} \right) = 1$$ for all $$\left( {x,y} \right)$$ in the region $$R,$$ the center of mass is defined only by the shape of the region and is called the centroid of $$R.$$

### Moments of Inertia of a Lamina

The moment of inertia of a lamina about the $$x$$-axis is defined by the formula
${{I_x} }={ \iint\limits_R {{y^2}\rho \left( {x,y} \right)dA} .}$ Similarly, the moment of inertia of a lamina about the $$y$$-axis is given by
${{I_y} }={ \iint\limits_R {{x^2}\rho \left( {x,y} \right)dA} .}$ The polar moment of inertia is
${{I_0} }={ \iint\limits_R {\left( {{x^2} + {y^2}} \right)\rho \left( {x,y} \right)dA} .}$

### Charge of a Plate

Suppose electrical charge is distributed over a region which has area $$R$$ in the $$xy$$-plane and its charge density is defined by the function $${\sigma \left( {x,y} \right)}.$$ Then the total charge $$Q$$ of the plate is defined by the expression
$Q = \iint\limits_R {\sigma \left( {x,y} \right)dA} .$

### Average of a Function

We give here the formula for calculation of the average value of a distributed function. Let $${f \left( {x,y} \right)}$$ be a continuous function over a closed region $$R$$ in the $$xy$$-plane. The average value $$\mu$$ of the function $${f \left( {x,y} \right)}$$ in the region $$R$$ is given by the formula
$\mu = \frac{1}{S}\iint\limits_R {f\left( {x,y} \right)dA} ,$ where $$S = \iint\limits_R {dA}$$ is the area of the region of integration $$R.$$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Find the centroid of the lamina cut by the parabolas $${y^2} = x$$ and $$y = {x^2}.$$

### ✓Example 2

Calculate the moments of inertia of the triangle bounded by the straight lines $$x + y = 1,$$ $$x = 0,$$ $$y = 0$$ (Figure $$2$$) and having density $$\rho \left( {x,y} \right) = xy.$$

### ✓Example 3

Electric charge is distributed over the disk $${x^2} + {y^2} = 1$$ so that its charge density is $$\sigma \left( {x,y} \right) =$$ $$1 + {x^2} + {y^2}$$ $$\left( {\text{Kl/m}^2} \right).$$ Calculate the total charge of the disk.

### Example 1.

Find the centroid of the lamina cut by the parabolas $${y^2} = x$$ and $$y = {x^2}.$$

#### Solution.

The lamina has the form shown in Figure $$1.$$ Since it is homogeneous, we suppose that the density $$\rho \left( {x,y} \right) = 1.$$ The mass of the lamina is
${m = \iint\limits_R {dA} } = {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]dx} } = {\int\limits_0^1 {\left[ {\left. y \right|_{{x^2}}^{\sqrt x }} \right]dx} } = {\int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)dx} } = {\int\limits_0^1 {\left( {{x^{\large\frac{1}{2}\normalsize}} – {x^2}} \right)dx} } = {\left. {\left( {\frac{{2{x^{\large\frac{3}{2}\normalsize}}}}{3} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 } = {\frac{2}{3} – \frac{1}{3} }={ \frac{1}{3}.}$

Figure 1.

Now we find the moment of the lamina about the $$x$$-axis and $$y$$-axis.
${{M_x} = \iint\limits_R {ydA} } = {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {ydy} } \right]dx} } = {\int\limits_0^1 {\left[ {\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_{{x^2}}^{\sqrt x }} \right]dx} } = {\frac{1}{2}\int\limits_0^1 {\left( {x – {x^4}} \right)dx} } = {\frac{1}{2}\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^5}}}{5}} \right)} \right|_0^1 } = {\frac{1}{2}\left( {\frac{1}{2} – \frac{1}{5}} \right) }={ \frac{3}{{20}},}$ ${{M_y} = \iint\limits_R {xdA} } = {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]xdx} } = {\int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)xdx} } = {\int\limits_0^1 {\left( {{x^{\large\frac{3}{2}\normalsize}} – {x^3}} \right)dx} } = {\left. {\left( {\frac{{2{x^{\large\frac{5}{2}\normalsize}}}}{5} – \frac{{{x^4}}}{4}} \right)} \right|_0^1 } = {\frac{2}{5} – \frac{1}{4} }={ \frac{3}{{20}}.}$ Thus, the coordinates of the center of mass are
${{\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} }={ \frac{9}{{20}},\;\;}}\kern-0.3pt {{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} }={ \frac{9}{{20}}.}}$

Page 1
Problem 1
Page 2
Problems 2-3