The moment of inertia (also called the second moment) is a physical quantity which measures the rotational inertia of an object.

The moment of inertia can be thought as the rotational analogue of mass in the linear motion.

The moment of inertia of a body is always defined about a rotation axis.

### Moment of Inertia of Point Masses

For a single mass, the moment of inertia is expressed as

\[I = m{r^2},\]

where \(m\) is the mass of the object, and \(r\) is the distance from the object to the axis of rotation.

If a system consists of \(n\) bodies, then the moment of inertia is given by

\[{I = {m_1}r_1^2 + {m_2}r_2^2 + \cdots + {m_n}r_n^2 }={ \sum\limits_{i = 1}^n {{m_i}r_i^2} ,}\]

where \({m_1},{m_2},\ldots,{m_n}\) are the masses of the bodies, \({r_1},{r_2},\ldots,{r_n}\) are their distances from the axis of rotation.

We can represent the last equation in the form

\[I = MR_g^2,\]

where \(M = \sum\limits_{i = 1}^n {{m_i}} ,\) and \({R_g}\) is called the radius of gyration.

It follows from the above equation that

\[{R_g} = \sqrt {\frac{I}{M}} .\]

When all masses \({m_i}\) are the same:

\[{m_1} = {m_2} = \cdots = {m_n} = m,\]

then the moment of inertia can be written in the form

\[{I = m\left( {r_1^2 + r_2^2 + \cdots + r_n^2} \right) }={ m\sum\limits_{i = 1}^n {r_i^2} .}\]

In this case,

\[{{R_g} = \frac{1}{n}\sqrt {r_1^2 + r_2^2 + \cdots + r_n^2} }={ \frac{1}{n}\sqrt {\sum\limits_{i = 1}^n {r_i^2} } ,}\]

where \(n\) is the number of bodies in the system.

### Moment of Inertia of a Lamina

When we deal with distributed objects like a lamina, or a solid, we need to calculate the contribution of each infinitesimally small piece of mass \(dm\) to the total moment of inertia \(I.\) This can be done through integration. In general case, finding the moment of inertia requires double integration or triple integration. However, in some special cases, the problem can be solved using single integrals.

#### Case \(1.\) Density Depends on the \(x-\)Coordinate

Let a planar lamina be bounded by the curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) on the interval \(\left[ {a,b} \right].\) Suppose that the lamina is rotated about the \(y-\)axis.

If the density \(\rho\) only depends on the \(x-\)coordinate, then the moment of inertia of a thin rectangle of width \(dx\) is defined by the formula

\[{dI = {x^2}dm }={ {x^2}\rho \left( x \right)\left[ {f\left( x \right) – g\left( x \right)} \right]dx.}\]

The total moment of inertia of the lamina about the \(y-\)axis is given by the integral

\[{I \text{ = }}\kern0pt{\int\limits_a^b {{x^2}\rho \left( x \right)\left[ {f\left( x \right) – g\left( x \right)} \right]dx} .}\]

#### Case \(2.\) Density Depends on the \(y-\)Coordinate

Similarly, we can consider a region of type \(II,\) bounded by the curves \(x = f\left( y \right),\) \(x = g\left( y \right)\) and the horizontal lines \(y = c,\) \(y = d.\) If the density of such a region only depends on the variable \(y,\) that is \(\rho = \rho \left( y \right),\) then the moment of inertia \(I\) of the lamina can be expressed by the single integral

\[{I \text{ = }}\kern0pt{\int\limits_c^d {{y^2}\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy} .}\]

### Parallel Axis Theorem

Suppose that an object is rotated about an axis passing through the center of gravity of the object and has the moment of inertia \({I_C}.\) Then the moment of inertia \(I\) about any other axis of rotation, which is parallel to the initial axis is given by the parallel axis theorem (also known as Huygens–Steiner theorem):

\[I = {I_C} + m{d^2},\]

where \(m\) is the mass of the object, and \(d\) is the distance between the two axes.

By definition, the distance \(d\) is the perpendicular distance between the axes.

## Solved Problems

Click a problem to see the solution.

### Example 1

Find the moment of inertia of a rectangle with sides \(a\) and \(b\) with respect to an axis passing through the side \(b.\)### Example 2

Find the moment of inertia of the semi-circular arc of radius \(R\) and mass \(m\) about an axis passing through its diameter.### Example 3

Find the moment of inertia of a uniform thin disk of radius \(R\) and mass \(m\) rotating about an axis passing through its center.### Example 4

A thin uniform rod of length \(\ell\) and mass \(m\) is rotated about the axis which is perpendicular to the rod and passes through its end. Calculate the moment of inertia of the rod.### Example 5

Find the moment of inertia of a right circular cone of mass \(m,\) base radius \(R\) and height \(H\) with respect to its centroidal axis.### Example 6

A right uniform triangle with legs \(a\) and \(b\) and mass \(m\) is rotated about the leg \(b.\) Calculate the moment of inertia of the triangle.### Example 7

A uniform lamina is bounded by the curves \(y = \sqrt x\) and \(y = {x^2}\) and has the mass \(m.\) Find the moment of inertia of the lamina about the \(x-\)axis.### Example 8

Find the moment of inertia of a uniform ball of mass \(m\) and radius \(R\) with respect to a diameter.### Example 9

A uniform lamina of mass \(m\) is bounded by the exponential curve \(y = \exp \left({-x}\right)\) and the coordinate axes. Find the moment of inertia of the lamina about the \(y-\)axis.### Example 10

Given an ellipse with semi-major axis \(a\) and semi-minor axis \(b.\) Find the moment of inertia of the ellipse about the \(y-\)axis.### Example 1.

Find the moment of inertia of a rectangle with sides \(a\) and \(b\) with respect to an axis passing through the side \(b.\)Solution.

Consider a small strip of the rectangle of width \(dx.\) The distance of the strip from the axis of rotation is equal to \(x.\) Therefore, it has the moment of inertia

\[dI = {x^2}dm = {x^2}\rho bdx.\]

Assuming the density is \(\rho = 1,\) we can write

\[dI = b{x^2}dx.\]

Integrating from \(x = 0\) to \(x = a\) yields:

\[{I = \int\limits_0^a {b{x^2}dx} }={ b\int\limits_0^a {{x^2}dx} }={ \left. {\frac{{b{x^3}}}{3}} \right|_0^a }={ \frac{{b{a^3}}}{3}.}\]

### Example 2.

Find the moment of inertia of the semi-circular arc of radius \(R\) and mass \(m\) about an axis passing through its diameter.Solution.

We take the radius vector that forms an angle \(\theta\) with the positive direction of the \(x-\)axis and consider an infinitely small element \(d\ell\) of the arc which is determined by increment \(d\theta.\) The mass of the element \(d\ell\) is

\[dm = \frac{{d\ell}}{\ell}m = \frac{m}{{\pi R}}d\ell.\]

The moment of inertia of the element \(dm\) of the arc about the \(y-\)axis is given by

\[\require{cancel}{dI = {x^2}dm }={ \frac{{m{R^\cancel{2}}{{\cos }^2}\theta }}{{\pi \cancel{R}}}d\ell }={ \frac{{mR\,{{\cos }^2}\theta }}{\pi }d\ell.}\]

Recall that \(d\ell = Rd\theta.\) Then

\[dI = \frac{{m{R^2}{{\cos }^2}\theta }}{\pi }d\theta .\]

To calculate the total moment of inertia of the semi-circular arc, we integrate from \(\theta = – \large{\frac{\pi }{2}}\normalsize\) to \(\theta = \large{\frac{\pi }{2}}\normalsize:\)

\[{I = \int\limits_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{m{R^2}{{\cos }^2}\theta }}{\pi }d\theta } }={ \frac{{m{R^2}}}{\pi }\int\limits_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^2}\theta d\theta } }={ \frac{{m{R^2}}}{{2\pi }}\int\limits_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {1 + \cos 2\theta } \right)d\theta } }={ \frac{{m{R^2}}}{{2\pi }}\left. {\left( {\theta + \frac{{\sin 2\theta }}{2}} \right)} \right|_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} }={\frac{{m{R^2}}}{{2\pi }}\left[ {\frac{\pi }{2} – \left( { – \frac{\pi }{2}} \right)} \right]}={ \frac{{m{R^2}}}{2}.}\]

### Example 3.

Find the moment of inertia of a uniform thin disk of radius \(R\) and mass \(m\) rotating about an axis passing through its center.Solution.

Take an arbitrary thin ring of radius \(0 \lt r \lt R\) and thickness \(dr.\) The mass of the elementary ring is

\[dm = \rho dA = 2\pi\rho rdr,\]

where \(\rho\) is the density of the disk material, and \(dA\) is the area of the ring.

The moment of inertia of the ring is given by

\[dI = {r^2}dm = 2\pi \rho {r^3}dr.\]

To find the moment of inertia of the entire disk, we integrate from \(r = 0\) to \(r = R:\)

\[{I = 2\pi \rho \int\limits_0^R {{r^3}dr} }={ 2\pi \rho \left. {\frac{{{r^4}}}{4}} \right|_0^R }={ \frac{{\pi \rho {R^4}}}{2}.}\]

Note that the mass of the disk is

\[m = \rho A = \pi \rho {R^2},\]

so

\[I = \pi \rho {R^2} \times \frac{{{R^2}}}{2} = \frac{{m{R^2}}}{2}.\]

### Example 4.

A thin uniform rod of length \(\ell\) and mass \(m\) is rotated about the axis which is perpendicular to the rod and passes through its end. Calculate the moment of inertia of the rod.Solution.

First we determine the moment of inertia of the rod about the axis passing through the center of gravity.

Consider a small portion of the rod located at distance \(x\) from the center. If the width of the element is \(dx,\) then the moment of inertia of the element about the center is

\[d{I_C} = \rho {x^2}dx,\]

where \(\rho\) is the linear density of the rod.

The total moment of inertia is defined through integration:

\[{{I_C} = \int\limits_{ – \frac{l}{2}}^{\frac{l}{2}} {\rho {x^2}dx} }={ \left. {\frac{{\rho {x^3}}}{3}} \right|_{ – \frac{l}{2}}^{\frac{l}{2}} }={ \frac{\rho }{3}\left[ {\frac{{{l^3}}}{8} – \left( { – \frac{{{l^3}}}{8}} \right)} \right] }={ \frac{{\rho {l^3}}}{{12}}.}\]

Since the mass of the rod is

\[m = \rho l,\]

we have

\[{{I_C} = \frac{{\rho {l^3}}}{{12}} }={ \rho l \times \frac{{{l^2}}}{{12}} }={ \frac{{m{l^2}}}{{12}}.}\]

Suppose now that the rod is rotated about the axis passing through one of the ends.

The new axis is located at a distance \(d = \large{\frac{l}{2}}\normalsize\) from the center of the rod. Using the parallel axis theorem, we get

\[{I = {I_C} + m{d^2} }={ \frac{{m{l^2}}}{{12}} + m{\left( {\frac{l}{2}} \right)^2} }={ \frac{{m{l^2}}}{{12}} + \frac{{m{l^2}}}{4} }={ \frac{{m{l^2}}}{3}.}\]

### Example 5.

Find the moment of inertia of a right circular cone of mass \(m,\) base radius \(R\) and height \(H\) with respect to its centroidal axis.Solution.

We assume that the cone is a uniform solid, so its density is

\[{\rho = \frac{m}{V} = \frac{m}{{\frac{1}{3}\pi {R^2}H}} }={ \frac{{3m}}{{\pi {R^2}H}}.}\]

Consider a thin slice of thickness \(dz\) at height \(z.\) The mass of the slice is

\[dm = \pi\rho {r^2}dz.\]

By triangle similarity,

\[\frac{r}{{H – z}} = \frac{R}{H},\]

so

\[r = \frac{{R\left( {H – z} \right)}}{H} = R\left( {1 – \frac{z}{H}} \right),\]

and

\[{dm = \pi\rho {r^2}dz }={\pi \rho {R^2}{\left( {1 – \frac{z}{H}} \right)^2}dz.}\]

The moment of inertia of the slice is

\[{dI = \frac{{{r^2}dm}}{2} }={ \frac{{\pi\rho {R^4}{{\left( {1 – \frac{z}{H}} \right)}^4}dz}}{2} }={ \frac{{\pi\rho {R^4}{{\left( {H – z} \right)}^4}}}{{2{H^4}}}dz.}\]

Integrating from \(z = 0\) to \(z = H\) gives the moment of inertia of the cone:

\[{I = \int\limits_0^H {\frac{{\pi\rho {R^4}{{\left( {H – z} \right)}^4}}}{{2{H^4}}}dz} }={ \frac{{\pi\rho {R^4}}}{{2{H^4}}}\int\limits_0^H {{{\left( {H – z} \right)}^4}dz} .}\]

Make the substitution:

\[{H – z = u,}\;\; \Rightarrow {z = H – u,\;\;}\kern0pt{dz = – du.}\]

When \(z = 0,\) \(u = H,\) and when \(z = H,\) \(u = 0.\) Hence,

\[{\int\limits_0^H {{{\left( {H – z} \right)}^4}dz} }={ \int\limits_H^0 {{u^4}\left( { – du} \right)} }={ \int\limits_0^H {{u^4}du} }={ \left. {\frac{{{u^5}}}{5}} \right|_0^H }={ \frac{{{H^5}}}{5}.}\]

Then the moment of inertia is written in the form

\[{I = \frac{{\pi\rho {R^4}}}{{2{H^4}}} \times \frac{{{H^5}}}{5} }={ \frac{{\pi\rho {R^4}H}}{{10}}.}\]

Substitute the expression for \(\rho:\)

\[{I = \frac{{3m}}{{\pi {R^2}H}} \times \frac{{\pi {R^4}H}}{{10}} }={ \frac{{3m{R^2}}}{{10}}.}\]

### Example 6.

A right uniform triangle with legs \(a\) and \(b\) and mass \(m\) is rotated about the leg \(b.\) Calculate the moment of inertia of the triangle.Solution.

The moment of inertia of the triangular lamina about the \(y-\)axis is given by the integral

\[{I_y} = \int\limits_0^a {{x^2}\rho \left( x \right)f\left( x \right)dx} .\]

The function \(f\left( x \right)\) is the hypotenuse \(AB\) of the triangle. We can easily derive its equation using the two-point form:

\[{\frac{{x – {x_A}}}{{{x_B} – {x_A}}} = \frac{{y – {y_A}}}{{{y_B} – {y_A}}},}\;\; \Rightarrow {\frac{{x – a}}{{0 – a}} = \frac{{y – 0}}{{b – 0}},}\;\; \Rightarrow {1 – \frac{x}{a} = \frac{y}{b},}\;\; \Rightarrow {y = f\left( x \right) = b – \frac{b}{a}x.}\]

Assuming that the density \(\rho\) is constant, we have

\[{I = {I_y} }= {\rho \int\limits_0^a {{x^2}\left( {b – \frac{b}{a}x} \right)dx} }={ \rho \int\limits_0^a {\left( {b{x^2} – \frac{b}{a}{x^3}} \right)dx} }={ \rho \left. {\left( {\frac{{b{x^3}}}{3} – \frac{{b{x^4}}}{{4a}}} \right)} \right|_0^a }={ \rho \left( {\frac{{b{a^3}}}{3} – \frac{{b{a^3}}}{4}} \right) }={ \frac{{\rho b{a^3}}}{{12}}.}\]

Let’s now recall that the mass of the triangular lamina is

\[{m = \rho A = \frac{{\rho ab}}{2}.}\]

Then the moment of inertia is expressed as

\[{I = \frac{{\rho b{a^3}}}{{12}} }={ \frac{{\rho ab}}{2} \times \frac{{{a^2}}}{6} }={ \frac{{m{a^2}}}{6}.}\]

### Example 7.

A uniform lamina is bounded by the curves \(y = \sqrt x\) and \(y = {x^2}\) and has the mass \(m.\) Find the moment of inertia of the lamina about the \(x-\)axis.Solution.

We can calculate the moment of inertia of the lamina by the formula

\[{I = {I_x} }={ \rho \int\limits_0^1 {{y^2}\left[ {f\left( y \right) – g\left( y \right)} \right]dy} .}\]

We rewrite the equations of both curves as functions of \(y:\)

\[{y = \sqrt x ,}\;\; \Rightarrow {x = g\left( y \right) = {y^2};}\]

\[{y = {x^2},}\;\; \Rightarrow {x = f\left( y \right) = \sqrt y .}\]

Integration gives:

\[{I = \rho \int\limits_0^1 {{y^2}\left( {\sqrt y – {y^2}} \right)dy} }={ \rho \int\limits_0^1 {\left( {{y^{\frac{5}{2}}} – {y^4}} \right)dy} }={ \rho \left. {\left( {\frac{{2{y^{\frac{7}{2}}}}}{7} – \frac{{{y^5}}}{5}} \right)} \right|_0^1 }={ \rho \left( {\frac{2}{7} – \frac{1}{5}} \right) }={ \frac{{3\rho }}{{35}}.}\]

Now let’s determine the mass of the lamina:

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

Hence, the moment of inertia is

\[{I = \frac{{3\rho }}{{35}} }={ \frac{\rho }{3} \times \frac{9}{{35}} }={ \frac{{9m}}{{35}}.}\]

### Example 8.

Find the moment of inertia of a uniform ball of mass \(m\) and radius \(R\) with respect to a diameter.Solution.

The density of the ball is

\[{\rho = \frac{m}{V} }={ \frac{m}{{\frac{4}{3}\pi {R^3}}} }={ \frac{{3m}}{{4\pi {R^3}}}.}\]

Consider a thin slice of thickness \(dz\) at height \(z.\) We can express its radius in terms of the coordinate \(z:\)

\[r = \sqrt {{R^2} – {z^2}} .\]

The mass of the elementary slice is

\[{dm = \pi \rho {r^2}dz }={ \pi \rho \left( {{R^2} – {z^2}} \right)dz.}\]

The moment of inertia of the disk is given by

\[{dI = \frac{{{r^2}dm}}{2} }={ \frac{{\left( {{R^2} – {z^2}} \right)dm}}{2} }={ \frac{{\pi \rho {{\left( {{R^2} – {z^2}} \right)}^2}}}{2}dz.}\]

To find the moment of inertia of the ball, we integrate from \(z = -R\) to \(z = R:\)

\[{I = \int\limits_{ – R}^R {\frac{{\pi \rho {{\left( {{R^2} – {z^2}} \right)}^2}}}{2}dz} }={ \frac{{\pi \rho }}{2}\int\limits_{ – R}^R {{{\left( {{R^2} – {z^2}} \right)}^2}dz} }={ \pi \rho \int\limits_0^R {{{\left( {{R^2} – {z^2}} \right)}^2}dz} }={ \pi \rho \int\limits_0^R {\left( {{R^4} – 2{R^2}{z^2} + {z^4}} \right)dz} }={ \pi \rho \left. {\left( {{R^4}z – \frac{{2{R^2}{z^3}}}{3} + \frac{{{z^5}}}{5}} \right)} \right|_0^R }={ \pi \rho {R^5}\left( {1 – \frac{2}{3} + \frac{1}{5}} \right) }={ \frac{{8\pi \rho {R^5}}}{{15}}.}\]

Substituting the equation for \(\rho,\) we have

\[{I = \frac{{8\pi {R^5}}}{{15}} \times \frac{{3m}}{{4\pi {R^3}}} }={ \frac{{2m{R^2}}}{5}.}\]

### Example 9.

A uniform lamina of mass \(m\) is bounded by the exponential curve \(y = \exp \left({-x}\right)\) and the coordinate axes. Find the moment of inertia of the lamina about the \(y-\)axis.Solution.

The moment of inertia of the infinite lamina about the \(y-\)axis is determined by the improper integral

\[{I = {I_y} }={ \int\limits_0^\infty {{x^2}\rho \left( x \right)f\left( x \right)dx} }={ \rho \int\limits_0^\infty {{x^2}{e^{ – x}}dx} .}\]

Integrating by parts twice, we can write the indefinite integral in the form:

\[{\int {{x^2}{e^{ – x}}dx} }={ \left[ {\begin{array}{*{20}{l}} {u = {x^2}}\\ {dv = {e^{ – x}}dx}\\ {du = 2xdx}\\ {v = – {e^{ – x}}} \end{array}} \right] }={ – {x^2}{e^{ – x}} – \int {\left( { – 2x{e^{ – x}}} \right)dx} }={ – {x^2}{e^{ – x}} + 2\int {x{e^{ – x}}dx} }={ \left[ {\begin{array}{*{20}{l}} {u = x}\\ {dv = {e^{ – x}}dx}\\ {du = dx}\\ {v = – {e^{ – x}}} \end{array}} \right] }={ – {x^2}{e^{ – x}} + 2\left[ { – x{e^{ – x}} – \int {\left( { – {e^{ – x}}} \right)dx} } \right] }={ – {x^2}{e^{ – x}} + 2\left[ { – x{e^{ – x}} + \int {{e^{ – x}}dx} } \right] }={ – {x^2}{e^{ – x}} + 2\left[ { – x{e^{ – x}} – {e^{ – x}}} \right] }={ – {x^2}{e^{ – x}} – 2x{e^{ – x}} – 2{e^{ – x}} }={ – {e^{ – x}}\left( {{x^2} + 2x + 2} \right).}\]

Returning back to the improper integral, we have

\[{I = \rho \int\limits_0^\infty {{x^2}{e^{ – x}}dx} }={ – \rho \lim \limits_{b \to \infty } \left. {\left[ {{e^{ – x}}\left( {{x^2} + 2x + 2} \right)} \right]} \right|_0^b }={ – \rho \lim \limits_{b \to \infty } \left[ {{e^{ – b}}\left( {{b^2} + 2b + 2} \right) – 2} \right] }={ – \rho \lim \limits_{b \to \infty } \left[ {\frac{{{b^2} + 2b + 2}}{{{e^b}}} – 2} \right] }={ 2\rho – \rho \lim \limits_{b \to \infty } \frac{{{b^2} + 2b + 2}}{{{e^b}}}.}\]

By L’Hopital’s Rule,

\[{\lim \limits_{b \to \infty } \frac{{{b^2} + 2b + 2}}{{{e^b}}} }={ \left[ {\frac{\infty }{\infty }} \right] }={ \lim \limits_{b \to \infty } \frac{{\left( {{b^2} + 2b + 2} \right)^\prime}}{{\left( {{e^b}} \right)^\prime}} }={ \lim \limits_{b \to \infty } \frac{{2b + 2}}{{{e^b}}} }={ \left[ {\frac{\infty }{\infty }} \right] }={ \lim \limits_{b \to \infty } \frac{{\left( {2b + 2} \right)^\prime}}{{\left( {{e^b}} \right)^\prime}} }={ \lim \limits_{b \to \infty } \frac{2}{{{e^b}}} }={ 0.}\]

Thus, we see that the moment of inertia of the infinite lamina has the finite value equal to

\[I = 2\rho .\]

We can express the moment of inertia of the lamina in terms of its mass \(m.\) Given that

\[{m = \rho A = \rho \int\limits_0^\infty {f\left( x \right)dx} }={ \rho \int\limits_0^\infty {{e^{ – x}}dx} }={ – \left. {\rho {e^{ – x}}} \right|_0^\infty }={ – \rho \lim \limits_{b \to \infty } \left. {\left[ {{e^{ – x}}} \right]} \right|_0^b }={ – \rho \lim \limits_{b \to \infty } \left[ {{e^{ – b}} – 1} \right] }={ \rho ,}\]

the final answer is written as

\[I = 2m.\]

### Example 10.

Given an ellipse with semi-major axis \(a\) and semi-minor axis \(b.\) Find the moment of inertia of the ellipse about the \(y-\)axis.Solution.

The moment moment of inertia of the ellipse about the \(y-\)axis is given by the integral

\[{{I_y} }={ \rho \int\limits_{ – a}^a {{x^2}\left[ {f\left( x \right) – g\left( x \right)} \right]dx} ,}\]

where \(\rho\) is the surface density of the ellipse, and

\[{f\left( x \right) = \frac{b}{a}\sqrt {{a^2} – {x^2}} ,\;\;}\kern0pt{g\left( x \right) = – \frac{b}{a}\sqrt {{a^2} – {x^2}}. }\]

This yields:

\[{I_y} = \frac{{2\rho b}}{a}\int\limits_{ – a}^a {{x^2}\sqrt {{a^2} – {x^2}} dx} .\]

As the function in the integrand is even, we can rewrite the integral in the form

\[{I_y} = \frac{{4b\rho }}{a}\int\limits_0^a {{x^2}\sqrt {{a^2} – {x^2}} dx} .\]

Now we make the substitution

\[{x = a\cos t,\;\;}\kern0pt{dx = – a\sin tdt.}\]

When \(x = 0,\) then \(t = \large{\frac{\pi }{2}}\normalsize,\) and when \(x = a,\) then \(t = 0.\) So we have

\[{{I_y} = \frac{{4\rho b{a^4}}}{a}\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}t\,{{\sin }^2}t\,dt} }={ \rho b{a^3}\int\limits_0^{\frac{\pi }{2}} {{{\left( {2\cos t\sin t} \right)}^2}dt} }={ \rho b{a^3}\int\limits_0^{\frac{\pi }{2}} {{{\left( {\sin 2t} \right)}^2}dt} }={ \frac{{\rho b{a^3}}}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 – \cos 4t} \right)dt} }={ \frac{{\rho b{a^3}}}{2}\left. {\left[ {t – \frac{{\sin 4t}}{4}} \right]} \right|_0^{\frac{\pi }{2}} }={ \frac{{\rho b{a^3}}}{2} \times \frac{\pi }{2} }={ \frac{{\pi \rho b{a^3}}}{4}.}\]

Assuming that the density is \(\rho,\) we find the mass of the ellipse:

\[m = \rho A = \pi \rho ab,\]

where \(A\) denotes the area of the ellipse.

Hence,

\[{{I_y} = \frac{{\pi \rho b{a^3}}}{4} }={ \pi \rho ab \times \frac{{{a^2}}}{4} }={ \frac{{m{a^2}}}{4}.}\]

Notice that changing \(b \to a,\) \(a \to b\) gives us the moment of inertia of the ellipse about the \(x-\)axis:

\[{I_x} = \frac{{\pi \rho a{b^3}}}{4} = \frac{{m{b^2}}}{4}.\]