I have the following equation for the torque exerted on a turbine blade.
The paper that I am reading says, "the corresponding torque is given by..."
$$dT = ρV_2ωr^22πrdr$$
My problem is that while I do know that $d$ is the derivative operator in this case, how do I apply it? Why is the corresponding torque given by $dT$ as opposed to $T$? I am trying to find an expression for finding $T$ based on the variables needed.
I realize this may be simple, however, I ask because every google search I did yielded $\frac{dM}{dT}$ or something of the sort, not a simple $dM$.
My background: I am in high school and I have taught myself parts of differential and integral calculus. I am going to learn differential equations next, I would just like to know how to apply this operator.
Good day!
This expression $dT$ isn't an operator, it's a differential form, specifically a 1-form. Loosely speaking, it's a quantity that can be integrated over one variable and represents a differential (read: "infinitesimally small piece of...") torque at a point. Divide by $dr$ on both sides and you get $dT/dr$, the total derivative of torque with respect to the radial coordinate. Integrate with respect to $r$ over the volume/surface/whatever domain $\Omega$ (in your case the turbine blade, I believe) to get the total torque, i.e. $$T = \int_{\Omega} dT = \int_{\Omega} \dfrac{dT}{dr} dr$$ Make sure you fully grasp the (definite) integral as an infinite sum of infinitesimally small "pieces," and this should start to make a whole lot of sense.
Technically speaking, the operator $d$ is the exterior derivative, which corresponds with the total differential in real vector spaces.