I hope this question will be allowed on math.stackexchange, as the question is a mathematical one even though the subject might be from engineering.
I am trying to derive the formula for the second moment of area for a circle:
Definition of second moment of area about x axis: $I_x=\int_A y^2 dA$
"Translation" into polar coordinates yields $y=r\sin\theta$
What I do not understand is why an infinitesimally small area of a circle is $dA=rdrd\theta$, and not just $dA=drd\theta$.
Where is the extra $r$ from?
$\endgroup$2 Answers
$\begingroup$See the following picture:
The area $dA$ is approximated by the area of a rectangle. The length is $dr$, the width is $rd\theta$, the arc length. Hence $rdrd\theta$.
$\endgroup$ 2 $\begingroup$Whenever you transform variables in a surface integral, you need to introduce a factor inside the integral called a Jacobian. You can think of this as the equivalent to applying the chain rule for changing more that one differential variable, i.e. $dx dy$ to $dr d\theta$ instead of just, e.g. $dx$ to $du$.
For a 1D integral, you would use the chain rule like such:
$$dx = \frac{dx}{du} du$$
For 2D integrals, this becomes:
$$dx dy = J dr d\theta$$
where $J$ is defined as the determinant of a 2x2 matrix:
$$J = \left| \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\\ \end{matrix} \right| $$
Hence the extra $r$ that appeared is from the Jacobian.
Similar matrices exist for 3 dimensions (perhaps higher dimensions too?).
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