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Let the general polar region $D$ be $$D=\{(\rho,\theta)\mid \alpha\leq\theta\leq\beta, \varphi_1(\theta)\leq \rho\leq\varphi_2(\theta)\}$$ and let $f(x,y)$ be a Riemann integrable function over $D$. Then the the double integral $\iint_{D}f(x,y)dxdy$ can be computed by $$\iint_{D}f(x,y)dxdy=\int_\alpha^\beta \left(\int_{\varphi_1(\theta)}^{\varphi_2(\theta)}f(\rho\cos\theta,\rho\sin\theta)\rho d\rho\right)d\theta.$$

My question is what is the geometric meaning of integral $$\int_{\varphi_1(\theta)}^{\varphi_2(\theta)}f(\rho\cos\theta,\rho\sin\theta)\rho d\rho ?$$

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2 Answers 2

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If $f(x,y) = 1$, it means the rate of change of area with respect to $\theta$, or $\frac{dA}{d\theta}$. This comes up for example in Kepler's Laws of Planetary motion, sweeping out equal areas in equal times, $\frac{dA}{dt}$ can be related if you know $\frac{d\theta}{dt}$.

In general it means the rate of change of the quantity getting integrated with respect to $\theta$. The meaning will depend on what sort of thing is actually being computed. If you are imagining a $3D$ shape where $f(x,y)$ represents the height, then the expression you asked about is the rate of change of volume with respect to theta, $\frac{dV}{d\theta}$.

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    $\begingroup$ Thank you for your explanation. $\frac{d\theta}{dt}$ means the angular velocity? $\endgroup$
    – HGF
    Commented Apr 12, 2021 at 23:33
  • $\begingroup$ @HGF yes exactly. $\endgroup$ Commented Apr 29, 2022 at 1:46
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A magic phrase that might be helpful here (or might be further confusing) is volume form. Just as the expression $dx\ dy$ expresses the area of an infinitesimally small rectangle in cartesian coordinates, the expression $\rho\ d\rho\ d\theta$ expresses the area of an infinitesimally small rectangle in polar coordinates. So what does this mean for the inner integral? Well, there are a couple of different ways of looking at it. One is that it's the usual area-under-the-curve, weighted by the relative 'length' of the infinitesimally small cross-section. It also has an interpretation from a physical perspective, though: it's the moment integral with respect to the central point (or axis) $\rho=0$

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