# Calculus

## Double Integrals # Physical Applications of Double Integrals

• ### 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 a problem to see the 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

${m = \iint\limits_R {dA} } = {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]dx} } = {\int\limits_0^1 {\left[ {\left. y \right|_{{x^2}}^{\sqrt x }} \right]dx} } = {\int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)dx} } = {\int\limits_0^1 {\left( {{x^{\large\frac{1}{2}\normalsize}} – {x^2}} \right)dx} } = {\left. {\left( {\frac{{2{x^{\large\frac{3}{2}\normalsize}}}}{3} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 } = {\frac{2}{3} – \frac{1}{3} }={ \frac{1}{3}.}$

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