I wanted to know the variance of radius "r" for this half annulus shape's ability to resist moment bending by calculating it's polar moment of inertia/second moment of area. But, I ended up obtaining a negative value in this approach.
Assumptions :
$I_z$ is the pole axis (out from the plane/page) for polar moment of inertia and also the origin.
Outer radius "R" taken constant,
$M_z$ is the moment acting along z-axis (out from the plane/page) in CCW direction.
Formula $$J = ∫_A r^2 dA$$ .... refer Page 8
Calculation
$ A = \cfrac{\pi}{2} (R^2 - r^2) $
Differentiating..... *1 ,
$ \cfrac {dA}{dr} = -\pi r$
$ dA = -\pi r dr$ ......*2
$J = \int r^2 (- \pi r dr)$
$J = \int - r^3 \pi dr$
$J = \pi \cfrac {-r^4}{4}|^R_r$
$J = \cfrac{-\pi}{4} [R^4 - r^4] $
As of this I wanted to ask three questions,
First, the $r^2$ on the formula is the pythagorean distance of the element dA (in case of a complete solid circe - please refer figure 11 and equation 12 on page 8 - http://web.mit.edu/course/3/3.11/www/modules/torsion.pdf). For my case would the integrand be $(R-r)^2$ ?
Second, I do understand that to take the differential area "dA", I can multiply two differential length $(dx, dy)$ which form that area $(dA=dxdy)$ and I feel that would introduce double integrals. But the aforementioned pdf does not take double integrals rather finds the elemental area dA by differentiating $\left(\cfrac {dA}{dr} \right)$, which I assume and have applied the same on "*1". Is this approach correct ?
Third, on "*2" I remember in school we utilised tons of substitutions in those cases the integrand was substituted as well which changed the differential and it's limits making some sense. But, isn't the whole $\left(\cfrac {d}{dr} \right)$ an operator and not an algebraic fraction i.e. dA divided by dr, right ?. is this mathematically justifiable to perform algebraic fraction just on the differential? Where did I go wrong?
Thank you for your time and support.