Calculating radius of gyration of plane figure about x-axis

[Evari5te]

Homework Statement

K.A Stroud Engineering Mathematics 8th edition Kindle version Chapter 2.21 Further problems #1
A plane figure is enclosed by the x-axis and the curve $y=a\sin x$ between $x=0$ and $x=\pi$

Show that the radius of gyration $k$, of the figure about the x-axis is given by $\frac{a.\sqrt[]{2}}{3}$

Relevant Equations

Radius of gyration, $k = \sqrt[]{\left(\frac{I}{A}\right)}$ where
$I$ is the Area Moment of Inertia given by integrating the square of the distance $y$ from the x-axis to each infinitesimal element of area, $dA$ of the figure, multiplied by the area element itself.

$I=\int_0^\pi y^2dA$

$\therefore I = \int_0^\pi a^2 \sin^2 x dA$

$dA = y.dx = a \sin x dx$

$\therefore I = \int_0^\pi a^3 \sin^3 x dx = a^3 \int_0^\pi \sin^2 x.\sin x dx$

The closest I can get is $k=\frac{\sqrt{2}a}{\sqrt{3}}$ as follows:

$I=\int_0^\pi y^2dA$

$\therefore I = \int_0^\pi a^2 \sin^2 x dA$

$dA = y.dx = a \sin x dx$

$\therefore I = \int_0^\pi a^3 \sin^3 x dx = a^3 \int_0^\pi \sin^2 x.\sin x dx$

Let
$u=\sin^2x$, $\therefore \frac{du}{dx} = 2\sin x \cos x$

$\frac{dv}{dx}=\sin x$
$\therefore v = \int \sin x dx =-\cos x +C$
Integrating by parts
$I=a^3 \left\{uv - \int_0^\pi v\frac{du}{dx} dx \right\}$
$I = a^3 \left\{ \left[\sin^2 x \left(- \cos x \right) \right]_0^\pi - \int_0^\pi \left(- \cos x \right).2\cos x.\sin x dx \right\}$
Because $\sin 0=0$ and $\sin \pi =0$, $\left[\sin^2 x \left(- \cos x \right) \right]_0^\pi = 0 - 0$

leaving
$I = a^3 \left\{ - \int_0^\pi \left(- \cos x \right).2\cos x.\sin x dx \right\}$
Taking the $-1$ multiplier from the $\left(- \cos x \right)$ term and $2$ multipler from the $2\cos x.\sin x$ term outside of the integral (so negating the minus) gives
$I=+2a^3 \int_0^\pi \left(\cos x \right).\cos x.\sin x dx = +2a^3 \int_0^\pi \cos^2 x.\sin x dx$
Using the identity $\cos^2 x = 1 - \sin^2 x$
$I =2a^3 \int_0^\pi \left(1-\sin^2 x \right).\sin x dx$
$I=2a^3 \left\{ \int_0^\pi \sin x dx - \int_0^\pi \sin^3 x dx \right\}$
$I= 2a^3 \left[-\cos x \right]_0^\pi - 2a^3 \int_0^\pi \sin^3 x dx$
The term $a^3 \int_0^\pi \sin^3 x dx$ on the right hand side is now the same as the original integral so
$I = 2a^3\left[-\cos x \right]_0^\pi - 2I$
$\therefore 3I = 2a^3\left[-\cos x \right]_0^\pi$

$\cos \pi = -1 \therefore -\cos \pi = 1$
$\cos 0 = 1 \therefore -\cos 0 = -1$

$\therefore 3I = 2a^3\left[1 - \left(-1\right) \right] = 2a^3\left(2\right) = 4a^3$
$\therefore I = \frac{4a^3}{3}$


$A$ is the area of the figure,

$A=\int_0^\pi a \sin x dx = a\left[-cos x\right]_0^\pi = 2a$

$k=\sqrt[]{\frac{I}{A}}=\sqrt{\frac{4a^3}{3.(2a)}} = \sqrt{\frac{2a^2}{3}}=\frac{\sqrt{2}a}{\sqrt{3}}$

Am I wrong or is there a typo in the book?

 

[PhDeezNutz]

I'm getting the same answer as the book. I disagree with your $I$

I agree with the integral you set up but I think you are going the wrong way about evaluating it. That is to say making it too complex/long and prone to errors. Try this

$I = \int_{0}^{\pi} \frac{a^3}{3} \sin^3 x \, dx = \int_{0}^{\pi} \frac{a^3}{3} \left(1 - \cos^2 x \right) \sin x \, dx$

make $u = \cos x$ and do you $u$-substitution and see if you get

$I = \frac{4a^3}{9}$

 

[PhDeezNutz]

Typo in last post fixed

 

[Evari5te]

I'm getting the same answer as the book. I disagree with your $I$

I agree with the integral you set up but I think you are going the wrong way about evaluating it. That is to say making it too complex/long and prone to errors. Try this

$I = \int_{0}^{\pi} \frac{a^3}{3} \sin^3 x \, dx = \int_{0}^{\pi} \frac{a^3}{3} \left(1 - \cos^2 x \right) \sin x \, dx$

make $u = \cos x$ and do you $u$-substitution and see if you get

$I = \frac{4a^3}{9}$

Thanks PhDeezNutz. There is a key difference in your initial integral to mine which is the $\frac{1}{3}$ multiplier.

This entirely accounts for the mistake in my attempt, at it remains a constant multiplier along with the $a^3$ throughout and outside of the work resolving the integral.

This results in $3x3=9$ in the last denominator for $I$, which then square roots to 3 in the $k=\sqrt{\frac{I}{A}}$ calc. Since I didn't include it in my initial integral I end up with $\sqrt{3}$ in my final answer for $k$.

It's not immediately clear to me why we divide by 3 if rotating the figure around the x-axis instead of the y axis. Perhaps I'm mixing examples using area strips with those using infinitesimal area elements, and getting it wrong at the very first stage.

Is $dA=dy.dx$ rather than $dA =y.dx$ as I initially set out in the relevant equations?
Then would
$I = \int_0^\pi y^2 dA = \int_0^\pi y^2 dy.dx$
and the $y^2.dy$ be integrated as a first step to give the $\frac{y^3}{3}$ term?

At least this gives the right answer, but I'm not very confident my rationale is correct...

 

[PhDeezNutz]

I believe your rationale is correct (it's the one I used)

$I = \int_{x=0}^{\pi} \int_{y=0}^{a \sin x} y^2 \, dy dx$

$I = \int_{x=0}^{\pi} \left.\frac{y^3}{3} \right|_{y=0}^{a\sin x} \, dx$

etc....That $\frac{1}{3}$ turns to $\frac{1}{9}$ later when you integrate another "$u^2$" as outlined in post 2

 

[erobz]

Am I wrong or is there a typo in the book?

What is wrong is that your differential element is incorrect for the application of the definition. It needs to be horizontal here such that the element has a singular value for $y$ at each differential strip. Slice it parallel to $x$ axis if you are going to try that type of maneuvering (it will be probably be a headache in comparison to the double integration).

in the left, if you take $y$ at the centroid of the element $y/2$, you'll notice you get a factor of a $\frac{1}{4} \int y^2 dA$, so that doesn't fix your problem with $I$ if its indeed to be a $\frac{1}{3}$ of your result.