How do we integrate with respect to (dx)^2 (not dx)?

[AakashPandita]

$$∫x(dx)^2=?$$

and what is the difference between dx^2 and (dx)^2?

 

[pwsnafu]

Where's this from? I've seen $dx^2$ but not $(dx)^2$. The only way I can parse the latter is with wedge products

 

[Simon Bridge]

The integral is zero.
i.e. it has no contribution to the overall sum.

But let's be careful: where have you seen this come up?

 

[jackmell]

Isn't that just short-hand notation for a double integral, i.e.,

$$\iint x dx dx$$

 

[pwsnafu]

Isn't that just short-hand notation for a double integral, i.e.,

$$\iint x dx dx$$

I don't think you are allowed to have the same variable for multiple integrals. A double integral is integrating with respect to the area element.

 

[jackmell]

I don't think you are allowed to have the same variable for multiple integrals. A double integral is integrating with respect to the area element.

Ok, but the reason I suggested this is because I recall in one of my books on integral equations, the author, when converting from IVP to Volterra Equations, use the syntax such as:

$$\int_0^x \int_0^x f(t) dt dt=\int_0^x (x-t) f(t)dt$$

however, to be fair, he never used the syntax $(dt)^2$ so if I am in error, my apologies.

 

[AakashPandita]

I was trying to find the moment of inertia of a disk rotating at the axis through its center and perpendicular to it.

i got an integral for some function with (dx)^2 in it.

 

[HallsofIvy]

Ok, but the reason I suggested this is because I recall in one of my books on integral equations, the author, when converting from IVP to Volterra Equations, use the syntax such as:

$$\int_0^x \int_0^x f(t) dt dt=\int_0^x (x-t) f(t)dt$$

I strongly suspect that you have misread
$$\int_0^x\int_0^x f(t)d\tau dt$$

however, to be fair, he never used the syntax $(dt)^2$ so if I am in error, my apologies.

Neither "$dtdt$" nor "$(dx)^2$" has any meaning.

 

[jackmell]

I strongly suspect that you have misread
$$\int_0^x\int_0^x f(t)d\tau dt$$

With all due respect Hall, I have not. It's from "A First Course in Integral Equations" by Abdul-Majid Wazwaz. It's on p. 20 and includes also, the expression:

$$\int_0^x \int_0^x \int_0^x f(t)dt dt dt=1/2 \int_0^x (x-t)^2 f(t)dt$$

 

[HallsofIvy]

Then it is just bad notation.

 

[qbert]

With all due respect Hall, I have not. It's from "A First Course in Integral Equations" by Abdul-Majid Wazwaz. It's on p. 20 and includes also, the expression:

$$\int_0^x \int_0^x \int_0^x f(t)dt dt dt=1/2 \int_0^x (x-t)^2 f(t)dt$$

Dear lord, that's awful. I'd read the lhs as
$$\int_0^x \int_0^x \int_0^x f(t)dt dt dt<br /> = \int_0^x \int_0^x \left( \int_0^x f(t)dt \right) dt_1 \; dt_2 <br /> = \int_0^x \int_0^x H(x) dt_1 \; dt_2 <br /> = x^2 H(x)<br /> = x^2 \int_0^x f(t) dt$$

while, what was probably meant is
$$\int_0^x \int_0^{t_2} \int_0^{t_1} f(t) dt \; dt_1 \; dt_2 = \frac{1}{2} \int_0^x (x-t)^2 f(t) dt.$$

 

[economicsnerd]

Yikes, that's awful notation.

 

[pwsnafu]

With all due respect Hall, I have not. It's from "A First Course in Integral Equations" by Abdul-Majid Wazwaz. It's on p. 20 and includes also, the expression:

$$\int_0^x \int_0^x \int_0^x f(t)dt dt dt=1/2 \int_0^x (x-t)^2 f(t)dt$$

That's the Cauchy formula for repeated integration. And you are not allowed to write it as dtdtdt. You must keep the variables separate.
$\int_{a}^x \int_a^{t_1}\ldots \int_a^{t_{n-1}}f(t_n) \, dt_n \, dt_{n-1} \ldots dt_1 = \frac{1}{(n-1)!} \int_a^x (x-t)^{n-1} f(t) \, dt.$

 

[jackmell]

That's the Cauchy formula for repeated integration. And you are not allowed to write it as dtdtdt. You must keep the variables separate.
$\int_{a}^x \int_a^{t_1}\ldots \int_a^{t_{n-1}}f(t_n) \, dt_n \, dt_{n-1} \ldots dt_1 = \frac{1}{(n-1)!} \int_a^x (x-t)f(t) \, dt.$

Ok. Thanks! That makes sense to me now. Sorry for causing trouble.

 

[Simon Bridge]

I was trying to find the moment of inertia of a disk rotating at the axis through its center and perpendicular to it.

i got an integral for some function with (dx)^2 in it.

Please show how you derived that.

 

[AakashPandita]

$$I=\int_0^R dmr^2 dm=σπ[(r+dr)^2-r^2]$$
and then I put $$dm$$ in first.

 

[AakashPandita]

$$I=\int_0^R dmr^2 ;dm=σπ[(r+dr)^2-r^2]$$
and then I put $$dm$$ in first.

 

[pwsnafu]

What does $(r+dr)^2$ mean?

Edit: I think it should read $dm = 2\sigma \pi r \, dr$, which is compatible with what you wrote if we take $(dr)^2 = 0$.

Was there a textbook or a teacher who said you could square differentials? What made you think $(r+dr)^2$ is well defined?

 

[AakashPandita]

i tried to find the area of a ring by subtracting the area of smaller circle from larger circle.
The difference in the radii is dr.

area of ring $$=π[(r+dr)^2 - r^2]$$

 

[pwsnafu]

Gotcha. Yeah you use $(dr)^2 = 0$. See this thread.

 

[Simon Bridge]

$$I=\int_0^R dmr^2 dm=σπ[(r+dr)^2-r^2]$$
and then I put $$dm$$ in first.

Why do you have "dm" in twice?

This is using the concentric shells approach ... so the volume of the cylindrical shell between $r$ and $r+dr$ is the difference in the volumes of the cylinders:

$\pi h (r+dr)^2 - \pi h r^2$ where h is the height of the cylinder.

If you multiply that out, you get $\pi h (2rdr + dr^2)$ ... but $dr^2 = 0$ ... remember that "dr" comes from a limit?

You can think of it like this:
If $dr$ is infinitesimally small, then any large power of it must be even smaller than that.
But "infinitesimal" is the smallest you can get without actually being zero.
Therefore...

[edit]pwnsnafu beat me to it :)The volume of the shell between r and r+dr is much better derived by taking the surface area of the cylinder radius r (excluding the end-caps) and multiplying it by dr. You can work out why this works by sketching it out.