Calculating dA for Moment of Inertia of a Circle

[ShawnD]

The moment of inertia (I) for something is

$$I = \int y^2 dA$$

How do I get dA for a circle?
Here is what dA should look like for moment of inertia:
http://myfiles.dyndns.org/pictures/circle_dA.png

dA in this case is NOT a derivation of $$A = \pi r^2$$. Doing that will give you the polar moment of inertia (J) which is completely different.


If the integration was done correctly, the answer you get should be

$$I = \frac{\pi r^4}{4}$$

 

[Hurkyl]

$dA$ is an "element of area", much like $dx$ is an "element of length".

A two dimensional integral is computed by the limit of a two dimensional Riemann sum; but instead of partitioning an interval into arbitrarily small intervals, you partition your region into arbitrarily small rectangles.

In other words, an element of area is an infinitessimally small rectangle. So, it cannot span the entire width of your disk¹!


However, notice that the integrand is the same at every point along a horizontal line. So, adding up $y^2 dA$ for all of the rectangles in a row is the same as adding up $L y^2 dy$ where L is the width of the element you drew in your picture (and is a function of y), and that allows you to collapse the problem to a 1-d integral.


Or, if you don't like thinking ahead, you can simply brute force it:

$$\int_{-r}^r \int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}} y^2 \, dx \, dy$$


Hurkyl

1: Your shape is a disk, not a circle, because the interior is part of your shape.

 

[HallsofIvy]

Hurkyl was completely correct but I think what ShawnD meant to say was not "dA for a circle" but "dA in polar coordinates".

Doing problems with a circular symmetry, as in finding the area of a circle or finding the moment of inertia of a circle about a diameter, is most often easier in polar coordinates.

The "dA" in polar coordinates is "rdrdθ".

Finding the area of a circle of radius R is trivial in polar coordinates: it is $$\int_{r=0}^R\int_{\theta=0}^{2\pi}rdrd\theta= \{\int_{r=0}^R rdr\}\{\int_{\theta=0}^{2\pi}d\theta\}= 2\pi\(\frac{1}{2}R^2}= \pi R^2$$

The moment of inertia about the x-axis is
$$\int_{r=0}^R\int_{\theta=0}^{2\pi}y^2 rdrd\theta$$
and, since y= r sinθ, that is
$$\int_{r=0}^R\int_{\theta=0}^{2\pi}r^3 sin^2\theta drd\theta$$
$$=\{\int_{r=0}^R r^3 dr\}\{\int_{\theta=0}^{2\pi}sin^2\theta d\theta\}$$

 

[ShawnD]

Originally posted by Hurkyl
Or, if you don't like thinking ahead, you can simply brute force it:

$$\int_{-r}^r \int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}} y^2 \, dx \, dy$$

Can you explain a little more how you came up with this? I've only taken 1 calculus course ever and it didn't go into details about how double integrals work or how to just make equations.

Why does it have to be integrated between -r and r?

 

[Hurkyl]

What's the biggest and smallest y can be?
Once you've chosen y, what's the biggest and smallest x can be?

 

[ShawnD]

Originally posted by HallsofIvy

$$=\{\int_{r=0}^R r^3 dr\}\{\int_{\theta=0}^{2\pi}sin^2\theta d\theta\}$$

This one looks like it's getting very close... but how can I get a $\pi$ in there?

 

[da_willem]

sin(x)^2=.5-.5cos(2x), wanneer je dit naar x integreert krijg je .5x-.25sin(2x). Over de grenzen 0 tot 2pi geeft alleen de .5x term een bijdrage, en wel de benodigde pi.

 

[NateTG]

Originally posted by da_willem
sin(x)^2=.5-.5cos(2x), wanneer je dit naar x integreert krijg je .5x-.25sin(2x). Over de grenzen 0 tot 2pi geeft alleen de .5x term een bijdrage, en wel de benodigde pi.

Not that I can speak Dutch (or Flemmish) but the translation into English is probably:

$$\sin^2(x)=.5-.5cos(2x)$$, if I then integrate with respect to $$x$$ I get $$.5x -.25\sin(2x)$$. With the limits $$0$$ to $$2\pi$$ the $$.5x$$ term contributes the necessary $$\pi$$

 

[ShawnD]

Thanks everybody. It works now .