# Center of Mass and Moments

In this section we consider centers of mass and moments.

The center of mass for an object can be thought as the point about which the entire mass of the object is equally distributed.

If the object has a uniform density $$\rho,$$ then the center of mass is also the geometric center of the object called the centroid.

### Center of Mass and Moment of a Thin Rod

Suppose that we have a thin rod lying on the $$x-$$axis between $$x = a$$ and $$x = b.$$ At a point $$x,$$ the rod has mass density (mass per unit length) $$\rho \left( x \right).$$

The center of mass of the rod is given by

$\bar x = \frac{{{M_0}}}{m} = \frac{{\int\limits_a^b {x\rho \left( x \right)dx} }}{{\int\limits_a^b {\rho \left( x \right)dx} }}.$

The integral in the numerator $${{M_0} = \int\limits_a^b {x\rho \left( x \right)dx} }$$ is called the moment (or the first moment) of the one-dimensional object around zero.

The integral in the denominator $${m = \int\limits_a^b {\rho \left( x \right)dx} }$$ gives the total mass of the rod.

### Center of Mass and Moments of a Planar Lamina

In general, the center of mass and moments of a lamina can be determined using double integrals. However, in certain special cases when the density only depends on one coordinate, the calculations can be performed using single integrals.

#### Case $$1.$$ Density Depends on the $$x-$$Coordinate

Consider a thin planar lamina bounded by the curves $$y = f\left( x \right)$$ and $$y = g\left( x \right)$$ on the interval $$\left[ {a,b} \right].$$

If the density only depends on the $$x-$$coordinate, the moments of the lamina about the $$x-$$ and $$y-$$axes are

${M_x} = \frac{1}{2}\int\limits_a^b {\rho \left( x \right)\left[ {{f^2}\left( x \right) – {g^2}\left( x \right)} \right]dx} ,$

${M_y} = \int\limits_a^b {x\rho \left( x \right)\left[ {f\left( x \right) – g\left( x \right)} \right]dx} .$

The center of mass $$\left( {\bar x,\bar y} \right)$$ is given by

${\bar x = \frac{{{M_y}}}{m},\;\;}\kern0pt{\bar y = \frac{{{M_x}}}{m},}$

where $$m = \int\limits_a^b {\rho \left( x \right)\left[ {f\left( x \right) – g\left( x \right)} \right]dx}$$ is the mass of the lamina.

If the density $$\rho$$ is constant, these formulas are simplified. In this case the centroid of the lamina is determined by formulas

${\bar x = \frac{{{M_y}}}{m} }={ \frac{{\int\limits_a^b {x\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }}{{\int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }},}$

${\bar y = \frac{{{M_x}}}{m} }={ \frac{{\frac{1}{2}\int\limits_a^b {\left[ {{f^2}\left( x \right) – {g^2}\left( x \right)} \right]dx} }}{{\int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }}.}$

#### Case $$2.$$ Density Depends on the $$y-$$Coordinate

Similarly, we can consider a region enclosed by two curves $$x = f\left( y \right),$$ $$x = g\left( y \right)$$ and two horizontal lines $$y = c,$$ $$y = d.$$

Assuming the density of the region only depends on the $$y-$$coordinate, the moments of the lamina about the $$x-$$ and $$y-$$axes are expressed by the integrals

${M_x} = \int\limits_c^d {y\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy} ,$

${M_y} = \frac{1}{2}\int\limits_c^d {\rho \left( y \right)\left[ {{f^2}\left( y \right) – {g^2}\left( y \right)} \right]dy} .$

The center of mass is calculated by the formulas

${\bar x = \frac{{{M_y}}}{m},\;\;}\kern0pt{\bar y = \frac{{{M_x}}}{m},}$

where the mass of the lamina is now determined by the integral

$m = \int\limits_c^d {\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy} .$

In case of constant density, the centroid has the following coordinates

${\bar x = \frac{{{M_y}}}{m} }={ \frac{{\frac{1}{2}\int\limits_c^d {\left[ {{f^2}\left( y \right) – {g^2}\left( y \right)} \right]dy} }}{{\int\limits_c^d {\left[ {f\left( y \right) – g\left( y \right)} \right]dy} }},}$

${\bar y = \frac{{{M_x}}}{m} }={ \frac{{\int\limits_c^d {y\left[ {f\left( y \right) – g\left( y \right)} \right]dy} }}{{\int\limits_c^d {\left[ {f\left( y \right) – g\left( y \right)} \right]dy} }}.}$

### Centroid of a Planar Lamina in Polar Coordinates

Suppose that a polar region is bounded by two radius vectors $$\theta = \alpha,$$ $$\theta = \beta$$ and a continuous curve $$r = f\left( \theta \right).$$

The centroid $$\left( {\bar x,\bar y} \right)$$ of the polar region is given by the formulas

${\bar x = \frac{2}{3}\frac{{\int\limits_\alpha ^\beta {{r^3}\left( \theta \right)\cos \theta d\theta } }}{{\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }},\;\;}\kern0pt{\bar y = \frac{2}{3}\frac{{\int\limits_\alpha ^\beta {{r^3}\left( \theta \right)\sin \theta d\theta } }}{{\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }}.}$

### Centroid of a Triangle

The centroid of a triangle is the point of intersection of the medians of the triangle.

The centroid $$G\left( {\bar x,\bar y} \right)$$ of a triangle with vertices $$\left( {{x_1},{y_1}} \right),$$ $$\left( {{x_2},{y_2}} \right),$$ $$\left( {{x_1},{y_1}} \right)$$ has the coordinates

${\bar x = \frac{{{x_1} + {x_2} + {x_3}}}{3},\;\;}\kern0pt{\bar y = \frac{{{y_1} + {y_2} + {y_3}}}{3}.}$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A rod has a length of $$40\,\text{cm}.$$ The rod’s density changes linearly along its length from $$20\large{\frac{\text{g}}{\text{cm}}}\normalsize$$ to $$60\large{\frac{\text{g}}{\text{cm}}}\normalsize.$$ Find the center of mass of the rod.

### Example 2

Find the centroid of a semicircular region of radius $$1$$ centered at the origin.

### Example 3

Determine the center of mass of the lamina of uniform density bounded by the sine curve $$y = \sin x$$ and the $$x-$$axis on the interval $$\left[ {0,\pi } \right].$$

### Example 4

A square with side length $$a$$ lies in the first quadrant with one vertex at the origin. Its sides are parallel to the coordinate axes. Find the center of mass of the square if the density distribution is given by the linear function $$\rho \left( x \right) = kx.$$

### Example 5

Find the centroid of the region bounded by the cubic curve $$y = x^3,$$ the vertical line $$x = 1,$$ and the $$x-$$axis.

### Example 6

Find the centroid of the region enclosed by the curves $$y = \sqrt x$$ and $$y = {x^2}.$$

### Example 7

Find the centroid of the region bounded by the arc of the ellipse $$\large{\frac{{{x^2}}}{{{a^2}}}}\normalsize + \large{\frac{{{y^2}}}{{{b^2}}}}\normalsize = 1$$ lying in the first quadrant and the coordinate axes.

### Example 8

An isosceles triangle with vertices $$A( – 1,0),$$ $$B(1,0),$$ $$C(0,2)$$ has the density distribution according to the law $$\rho \left( y \right) = 1 + {y^2}.$$ Find the center of mass of the triangle.

### Example 9

Find the centroid of a circular sector with radius $$R$$ and angle $$\alpha.$$

### Example 10

Find the centroid of the region enclosed by the cardioid $$r\left( \theta \right) = 1 + \cos \theta .$$

### Example 1.

A rod has a length of $$40\,\text{cm}.$$ The rod’s density changes linearly along its length from $$20\large{\frac{\text{g}}{\text{cm}}}\normalsize$$ to $$60\large{\frac{\text{g}}{\text{cm}}}\normalsize.$$ Find the center of mass of the rod.

Solution.

Let’s first derive the density distribution function. The equation of a straight line passing through the points $$\rho \left( 0 \right) = 20,$$ $$\rho \left( 40 \right) = 60$$ is given by

${\frac{{\rho – \rho \left( 0 \right)}}{{\rho \left( {40} \right) – \rho \left( 0 \right)}} = \frac{{x – 0}}{{40 – 0}},}\;\; \Rightarrow {\frac{{\rho – 20}}{{60 – 20}} = \frac{x}{{40}},}\;\; \Rightarrow {\rho = x + 20,}$

where $$\rho$$ is measured in $$\large{\frac{\text{g}}{\text{cm}}}\normalsize,$$ and $$x$$ is measured in $$\text{cm}.$$

Calculate the mass $$m$$ and the first moment $${M_0}$$ of the rod:

${m = \int\limits_a^b {\rho \left( x \right)dx} }={ \int\limits_0^{40} {\left( {x + 20} \right)dx} }={ \left. {\left( {\frac{{{x^2}}}{2} + 20x} \right)} \right|_0^{40} }={800 + 800}={ 1600\,\text{g}.}$

${{M_0} = \int\limits_a^b {x\rho \left( x \right)dx} }={ \int\limits_0^{40} {x\left( {x + 20} \right)dx} }={ \int\limits_0^{40} {\left( {{x^2} + 20x} \right)dx} }={ \left. {\left( {\frac{{{x^3}}}{3} + 10{x^2}} \right)} \right|_0^{40} }={ \frac{{64000}}{3} + 16000 \approx 37333\,\text{g}\kern-0.3pt\cdot\kern-0.3pt\text{cm}.}$

Hence, the center of mass $$G\left( {\bar x} \right)$$ is located at the point

$\bar x = \frac{{{M_0}}}{m} = \frac{{37333}}{{1600}} \approx 23.3\,\text{cm}$

### Example 2.

Find the centroid of a semicircular region of radius $$1$$ centered at the origin.

Solution.

It is obvious that the centroid $$G\left( {\bar x,\bar y} \right)$$ lies on the $$y-$$axis, so $$\bar x = 0.$$ To find the $$\bar y-$$coordinate, we need to compute the first moment $${M_x}.$$ Assuming $$\rho = 1,$$ we get

${{M_x} = \frac{1}{2}\int\limits_a^b {{f^2}\left( x \right)dx} }={ \frac{1}{2}\int\limits_{ – 1}^1 {\left( {1 – {x^2}} \right)dx} }={ \frac{1}{2}\left. {\left( {x – \frac{{{x^3}}}{3}} \right)} \right|_{ – 1}^1 }={ \frac{1}{2} \cdot \frac{4}{3} }={ \frac{2}{3}}$

Since $$\rho = 1,$$ the mass $$m$$ of the lamina is numerically equal to the area $$A$$ of the semicircle:

$m = A = \frac{{\pi {R^2}}}{2} = \frac{\pi }{2}.$

Then

$\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{2}{3}}}{{\frac{\pi }{2}}} = \frac{4}{{3\pi }}.$

Hence, the centroid is located at the point

$G\left( {\bar x,\bar y} \right) = G\left( {0,\frac{4}{{3\pi }}} \right).$

### Example 3.

Determine the center of mass of the lamina of uniform density bounded by the sine curve $$y = \sin x$$ and the $$x-$$axis on the interval $$\left[ {0,\pi } \right].$$

Solution.

We first find the mass of the lamina:

${m = \rho \int\limits_a^b {f\left( x \right)dx} }={ \rho \int\limits_0^\pi {\sin xdx} }={ \rho \left. {\left( { – \cos x} \right)} \right|_0^\pi }={ 2\rho ,}$

where $$\rho$$ denotes the density of the plate.

Calculate the moments $${M_x}$$ and $${M_y}:$$

${{M_x} = \frac{\rho }{2}\int\limits_a^b {\left[ {{f^2}\left( x \right) – {g^2}\left( x \right)} \right]dx} }={ \frac{\rho }{2}\int\limits_0^\pi {{{\sin }^2}xdx} }={ \frac{\rho }{4}\int\limits_0^\pi {\left( {1 – \cos 2x} \right)dx} }={ \frac{\rho }{4}\left. {\left( {x – \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi }={ \frac{{\pi \rho }}{4};}$

${{M_y} = \rho \int\limits_a^b {x\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }={ \rho \int\limits_0^\pi {x\sin xdx} }={ \left[ {\begin{array}{*{20}{l}} {du = \sin xdx}\\ {v = x}\\ {u = – \cos x}\\ {dv = 1} \end{array}} \right] }={ \rho \left[ {\left. {\left( { – x\cos x} \right)} \right|_0^\pi + \int\limits_0^\pi {\cos xdx} } \right] }={ \rho \left. {\left[ { – x\cos x + \sin x} \right]} \right|_0^\pi }={ \pi \rho .}$

Hence, the coordinates of the center of mass are

$\require{cancel}{\bar x = \frac{{{M_y}}}{m} = \frac{{\pi \cancel{\rho} }}{{2 \cancel{\rho} }} = \frac{\pi }{2};}$

$\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{\pi \bcancel{\rho} }}{4}}}{{2 \bcancel{\rho} }} = \frac{\pi }{8}.$

### Example 4.

A square with side length $$a$$ lies in the first quadrant with one vertex at the origin. Its sides are parallel to the coordinate axes. Find the center of mass of the square if the density distribution is given by the linear function $$\rho \left( x \right) = kx.$$

Solution.

Calculate the first moments of the square region about the coordinate axes. Here $$f\left( x \right) = a,$$ $$g\left( x \right) = 0,$$ so we have

${{M_x} = \frac{1}{2}\int\limits_0^a {\rho \left( x \right){f^2}\left( x \right)dx} }={ \frac{{k{a^2}}}{2}\int\limits_0^a {xdx} }={ \left. {\frac{{k{a^2}{x^2}}}{4}} \right|_0^a }={ \frac{{k{a^4}}}{4};}$

${{M_y} = \int\limits_0^a {x\rho \left( x \right)f\left( x \right)dx} }={ ka\int\limits_0^a {{x^2}dx} }={ \left. {\frac{{ka{x^3}}}{3}} \right|_0^a }={ \frac{{k{a^4}}}{3}.}$

The mass of the square region is given by

${m = \int\limits_0^a {\rho \left( x \right)f\left( x \right)dx} }={ ka\int\limits_0^a {xdx} }={ \left. {\frac{{ka{x^2}}}{2}} \right|_0^a }={ \frac{{k{a^3}}}{2}.}$

Hence, the center of mass is at the point $$G\left( {\bar x,\bar y} \right)$$ with coordinates

${\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{{k{a^4}}}{3}}}{{\frac{{k{a^3}}}{2}}} = \frac{{2a}}{3},\;\;}\kern0pt{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{k{a^4}}}{4}}}{{\frac{{k{a^3}}}{2}}} = \frac{a}{2}.}$

### Example 5.

Find the centroid of the region bounded by the cubic curve $$y = x^3,$$ the vertical line $$x = 1,$$ and the $$x-$$axis.

Solution.

Assuming that $$\rho = 1,$$ we calculate the first moments $${M_x}$$ and $${M_y}:$$

${{M_x} = \frac{1}{2}\int\limits_a^b {\rho \left( x \right)\left[ {{f^2}\left( x \right) – {g^2}\left( x \right)} \right]dx} }={ \frac{1}{2}\int\limits_0^1 {{{\left( {{x^3}} \right)}^2}dx} }={ \frac{1}{2}\int\limits_0^1 {{x^6}dx} }={ \left. {\frac{{{x^7}}}{{14}}} \right|_0^1 }={ \frac{1}{{14}};}$

${{M_y} = \int\limits_a^b {x\rho \left( x \right)\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }={ \int\limits_0^1 {\left( {x \cdot {x^3}} \right)dx} }={ \int\limits_0^1 {{x^4}dx} }={ \left. {\frac{{{x^5}}}{5}} \right|_0^1 }={ \frac{1}{5}.}$

The mass $$m$$ of the region is given by

${m = \int\limits_a^b {\rho \left( x \right)\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }={ \int\limits_0^1 {{x^3}dx} }={ \left. {\frac{{{x^4}}}{4}} \right|_0^1 }={ \frac{1}{4}.}$

It follows from here that

${\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{1}{5}}}{{\frac{1}{4}}} = \frac{4}{5},\;\;}\kern0pt{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{1}{{14}}}}{{\frac{1}{4}}} = \frac{2}{7}.}$

### Example 6.

Find the centroid of the region enclosed by the curves $$y = \sqrt x$$ and $$y = {x^2}.$$

Solution.

Let $$\rho$$ be the density of the lamina. The total mass is equal to

${m = \rho \int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }={ \rho \int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)dx} }={ \rho \left. {\left( {\frac{{2{x^{\frac{3}{2}}}}}{3} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 }={ \frac{\rho }{3}.}$

Compute the first moments $${M_x}$$ and $${M_y}:$$

${{M_x} = \frac{\rho }{2}\int\limits_a^b {\left[ {{f^2}\left( x \right) – {g^2}\left( x \right)} \right]dx} }={ \frac{\rho }{2}\int\limits_0^1 {\left[ {{{\left( {\sqrt x } \right)}^2} – {{\left( {{x^2}} \right)}^2}} \right]dx} }={ \frac{\rho }{2}\int\limits_0^1 {\left( {x – {x^4}} \right)dx} }={ \frac{\rho }{2}\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^5}}}{5}} \right)} \right|_0^1 }={ \frac{{3\rho }}{{20}};}$

${{M_y} = \rho \int\limits_a^b {x\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }={ \rho \int\limits_0^1 {x\left( {\sqrt x – {x^2}} \right)dx} }={ \rho \int\limits_0^1 {\left( {{x^{\frac{3}{2}}} – {x^3}} \right)dx} }={ \rho \left. {\left( {\frac{{2{x^{\frac{5}{2}}}}}{5} – \frac{{{x^4}}}{4}} \right)} \right|_0^1 }={ \frac{{3\rho }}{{20}}.}$

Hence, the centroid of the region $$G\left( {\bar x,\bar y} \right)$$ has the coordinates

$\require{cancel}{\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{{3 \bcancel{\rho} }}{{20}}}}{{\frac{ \bcancel{\rho} }{3}}} = \frac{9}{{20}},\;\;}\kern0pt{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{3 \bcancel{\rho} }}{{20}}}}{{\frac{ \bcancel{\rho} }{3}}} = \frac{9}{{20}}.}$

### Example 7.

Find the centroid of the region bounded by the arc of the ellipse $$\large{\frac{{{x^2}}}{{{a^2}}}}\normalsize + \large{\frac{{{y^2}}}{{{b^2}}}}\normalsize = 1$$ lying in the first quadrant and the coordinate axes.

Solution.

It is convenient to write the equation of the arc 0f the ellipse in explicit form:

${\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,}\;\; \Rightarrow {\frac{{{y^2}}}{{{b^2}}} = 1 – \frac{{{x^2}}}{{{a^2}}},}\;\; \Rightarrow {{y^2} = \frac{{{b^2}}}{{{a^2}}}\sqrt {{a^2} – {x^2}} ,}\;\; \Rightarrow {y = \frac{b}{a}\sqrt {{a^2} – {x^2}} .}$

Determine the mass of the region assuming that $$\rho = 1.$$

${m = \int\limits_0^a {f\left( x \right)dx} }={ \frac{b}{a}\int\limits_0^a {\sqrt {{a^2} – {x^2}} dx} .}$

This integral can be evaluated using the substitution

${x = a\sin t,}\;\; \Rightarrow {\sqrt {{a^2} – {x^2}} = a\cos t,\;\;}\kern0pt{dx = a\cos tdt.}$

When $$x = 0,$$ $$t = 0,$$ and when $$x = a,$$ $$t = \large{\frac{\pi}{2}}\normalsize.$$ So the mass of the region is given by

${m = \frac{b}{a}\int\limits_0^{\frac{\pi }{2}} {a\cos t \cdot a\cos tdt} }={ ab\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} }={ \frac{{ab}}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)dt} }={ \frac{{ab}}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}} }={ \frac{{ab}}{2} \cdot \frac{\pi }{2} }={ \frac{{\pi ab}}{4}.}$

Calculate the first moments $${M_x}$$ and $${M_y}.$$

${{M_x} = \frac{1}{2}\int\limits_0^a {{f^2}\left( x \right)dx} }={ \frac{{{b^2}}}{{2{a^2}}}\int\limits_0^a {\left( {{a^2} – {x^2}} \right)dx} }={ \frac{{{b^2}}}{{2{a^2}}}\left. {\left( {{a^2}x – \frac{{{x^3}}}{3}} \right)} \right|_0^a }={ \frac{{{b^2}}}{{2{a^2}}}\left( {{a^3} – \frac{{{a^3}}}{3}} \right) }={ \frac{{{b^2}a}}{3}.}$

${{M_y} = \int\limits_0^a {xf\left( x \right)dx} }={ \frac{b}{a}\int\limits_0^a {x\sqrt {{a^2} – {x^2}} dx} .}$

Make the substitution:

${{a^2} – {x^2} = {z^2},}\;\; \Rightarrow {- 2xdx = 2zdz,}\;\; \Rightarrow {xdx = – zdz.}$

When $$x = 0,$$ $$z = a,$$ and when $$x = a,$$ $$z = 0.$$ Hence,

${{M_y} = \frac{b}{a}\int\limits_a^0 {\left( { – {z^2}} \right)dz} }={ \frac{b}{a}\int\limits_0^a {{z^2}dz} }={ \left. {\frac{b}{a}\frac{{{z^3}}}{3}} \right|_0^a }={ \frac{{b{a^2}}}{3}.}$

The centroid of the region $$G\left( {\bar x,\bar y} \right)$$ has the following coordinates:

${\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{{b{a^2}}}{3}}}{{\frac{{\pi ab}}{4}}} = \frac{{4a}}{{3\pi }},\;\;}\kern0pt{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{{b^2}a}}{3}}}{{\frac{{\pi ab}}{4}}} = \frac{{4b}}{{3\pi }}.}$

### Example 8.

An isosceles triangle with vertices $$A( – 1,0),$$ $$B(1,0),$$ $$C(0,2)$$ has the density distribution according to the law $$\rho \left( y \right) = 1 + {y^2}.$$ Find the center of mass of the triangle.

Solution.

By symmetry, the center of mass $$G\left( {\bar x,\bar y} \right)$$ of the triangle must lie on the $$y-$$axis, so we need to determine only the $$\bar y-$$coordinate.

The density $$\rho$$ of the triangular lamina varies along the $$y-$$axis. Therefore, to calculate the first moment $${M_x},$$ we use the formula

${M_x} = \int\limits_c^d {y\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy}.$

The legs of the isosceles triangle have the following equations:

$AC:\;x = g\left( y \right) = \frac{y}{2} – 1;$

$BC:\;x = f\left( y \right) = -\frac{y}{2} + 1.$

Since

${f\left( y \right) – g\left( y \right) = \left( { – \frac{y}{2} + 1} \right) – \left( {\frac{y}{2} – 1} \right) }={ 2 – y,}$

the first moment $${M_x}$$ is equal to

${{M_x} = \int\limits_0^2 {y\left( {1 + {y^2}} \right)\left( {2 – y} \right)dy} }={ \int\limits_0^2 {\left( {2y – {y^2} + 2{y^3} – {y^4}} \right)dy} }={ \left. {\left( {{y^2} – \frac{{{y^3}}}{3} + \frac{{{y^4}}}{2} – \frac{{{y^5}}}{5}} \right)} \right|_0^2 }={ 4 – \frac{8}{3} + 8 – \frac{{32}}{5} }={ \frac{{44}}{{15}}.}$

Compute the mass of the lamina:

$\require{cancel}{m = \int\limits_c^d {\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy} }={ \int\limits_0^2 {\left( {1 + {y^2}} \right)\left( {2 – y} \right)dy} }={ \int\limits_0^2 {\left( {2 – y + 2{y^2} – {y^3}} \right)dy} }={ \left. {\left( {2y – \frac{{{y^2}}}{2} + \frac{{2{y^3}}}{3} – \frac{{{y^4}}}{4}} \right)} \right|_0^2 }={ \cancel{4} – 2 + \frac{{16}}{3} – \cancel{4} }={ \frac{{10}}{3}.}$

Hence,

$\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{44}}{{15}}}}{{\frac{{10}}{3}}} = \frac{{22}}{{25}}.$

Thus, the center of mass of the triangular lamina is at the point

$G\left( {\bar x,\bar y} \right) = G\left( {0,\frac{{22}}{{25}}} \right).$

### Example 9.

Find the centroid of a circular sector with radius $$R$$ and angle $$\alpha.$$

Solution.

Let $$G\left( {\bar x,\bar y} \right)$$ be the centroid of the circular sector. It is convenient to consider the circle as a polar curve given by the equation $$r\left( \theta \right) = R,$$ where the angle $$\theta$$ varies from $$0$$ to $$\alpha.$$

We can compute the coordinates $$\bar x,$$ $$\bar y$$ by the formulas

${\bar x = \frac{2}{3}\frac{{\int\limits_0^\alpha {{r^3}\left( \theta \right)\cos \theta d\theta } }}{{\int\limits_0^\alpha {{r^2}\left( \theta \right)d\theta } }},\;\;}\kern0pt{\bar y = \frac{2}{3}\frac{{\int\limits_0^\alpha {{r^3}\left( \theta \right)\sin \theta d\theta } }}{{\int\limits_0^\alpha {{r^2}\left( \theta \right)d\theta } }}.}$

This yields:

${\bar x = \frac{2}{3}\frac{{\int\limits_0^\alpha {{R^3}\cos \theta d\theta } }}{{\int\limits_0^\alpha {{R^2}d\theta } }} }={ \frac{{2R}}{3}\frac{{\int\limits_0^\alpha {\cos \theta d\theta } }}{{\int\limits_0^\alpha {d\theta } }} }={ \frac{{2R}}{3}\frac{{\left. {\left( {\sin \theta } \right)} \right|_0^\alpha }}{\alpha } }={ \frac{{2R\sin \alpha }}{{3\alpha }};}$

${\bar y = \frac{2}{3}\frac{{\int\limits_0^\alpha {{R^3}\sin \theta d\theta } }}{{\int\limits_0^\alpha {{R^2}d\theta } }} }={ \frac{{2R}}{3}\frac{{\int\limits_0^\alpha {\sin \theta d\theta } }}{{\int\limits_0^\alpha {d\theta } }} }={ \frac{{2R}}{3}\frac{{\left. {\left( { – \cos \theta } \right)} \right|_0^\alpha }}{\alpha } }={ \frac{{2R\left( {1 – \cos \alpha } \right)}}{{3\alpha }}.}$

Thus, the centroid of the circular sector is located at the point

${G\left( {\bar x,\bar y} \right) \text{ = }}\kern0pt{G\left( {\frac{{2R\sin \alpha }}{{3\alpha }},\frac{{2R\left( {1 – \cos \alpha } \right)}}{{3\alpha }}} \right).}$

### Example 10.

Find the centroid of the region enclosed by the cardioid $$r\left( \theta \right) = 1 + \cos \theta .$$

Solution.

Let the centroid be at the point $$G\left( {\bar x,\bar y} \right).$$ By symmetry, the $$\bar y$$ coordinate of the centroid is equal to zero. We calculate the $$\bar x$$ coordinate by the formula

$\bar x = \frac{2}{3}\frac{{\int\limits_\alpha ^\beta {{r^3}\left( \theta \right)\cos \theta d\theta } }}{{\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }}.$

The integral in the denominator $${\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }$$ is equal to $$2A,$$ where $$A$$ is the area of the polar region. We can easily evaluate it:

${\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }={ 2\int\limits_0^\pi {{{\left( {1 + \cos \theta } \right)}^2}d\theta } }={ 2\int\limits_0^\pi {\left( {1 + 2\cos \theta + {{\cos }^2}\theta } \right)d\theta } }={ 2\int\limits_0^\pi {\left( {\frac{3}{2} + 2\cos \theta + \frac{{\cos 2\theta }}{2}} \right)d\theta } }={ 2\left. {\left( {\frac{3}{2}\theta + 2\sin \theta + \frac{{\sin 2\theta }}{4}} \right)} \right|_0^\pi }={ 3\pi .}$

Consider now the integral in the numerator:

${\int\limits_\alpha ^\beta {{r^3}\left( \theta \right)\cos \theta d\theta } }={ \int\limits_\alpha ^\beta {{{\left( {1 + \cos \theta } \right)}^3}\cos \theta d\theta } .}$

The integrand is written in the form:

${{\left( {1 + \cos \theta } \right)^3}\cos \theta }={ \cos \theta + 3{\cos ^2}\theta + 3{\cos ^3}\theta + {\cos ^4}\theta .}$

Using the trig identities

${\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2},$

${\cos ^3}\theta = \frac{{3\cos \theta + \cos 3\theta }}{4},$

${\cos ^4}\theta = \frac{{\cos 4\theta + 4\cos 2\theta + 3}}{8},$

we can rewrite the integrand as follows:

${{\left( {1 + \cos \theta } \right)^3}\cos \theta }={ \frac{{15}}{8} + \frac{{13\cos \theta }}{4} + 2\cos 2\theta }+{ \frac{{3\cos 3\theta }}{4} + \frac{{\cos 4\theta }}{8}.}$

Integrating from $$\theta = 0$$ to $$\theta = 2\pi$$ gives the following answer:

${\int\limits_0^{2\pi } {{{\left( {1 + \cos \theta } \right)}^3}\cos \theta d\theta } }={ \frac{{15}}{8} \cdot 2\pi }={ \frac{{15\pi }}{4}}$

Hence, the $$\bar x$$ coordinate of the centroid is equal to

$\bar x = \frac{2}{3}\frac{{\frac{{15\pi }}{4}}}{{3\pi }} = \frac{5}{6},$

so the answer is $$G\left( {\bar x,\bar y} \right) = \left( {\large{\frac{5}{6}}\normalsize,0} \right).$$