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I tried proving the formula for area of circle, what I did was talk a circle of r radius and find the area under the curve that is $\int_{0}^{r}\int_{0}^{r} \sqrt{x^2+y^2} dxdy$ and then converting it to polar coordinate and solving it, which will give $\frac{1}{4}^{th}$ of the total area, why is this approach wrong?

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  • $\begingroup$ I think the approach is correct $\endgroup$ – vidyarthi Nov 26 '16 at 10:52
  • $\begingroup$ @vidyarthi But when we change to polar coordinated the equation changes to $\int_{0}^{\pi/2}\int_{0}^{r} r^2drd\theta$ but the real equation should be $\int_{0}^{\pi/2}\int_{0}^{r} rdrd\theta$ $\endgroup$ – Kinshuk Dua Nov 26 '16 at 10:57
  • $\begingroup$ Can you Explain more detail in why you choose $\sqrt{x^2+y^2}$ to integrate and how do you find your boundary? you might find the reason after thinking these. $\endgroup$ – Ching-Ting Wu Nov 26 '16 at 10:59
  • $\begingroup$ @KinshukDua thanks to Ching-Ting Wu, the integral should be $4\int_0^r\int_0^{\sqrt{r^2-y^2}}dx dy$ $\endgroup$ – vidyarthi Nov 26 '16 at 11:02
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    $\begingroup$ Because th equation $r=\sqrt{x^2+y^2}$ is Exactly your boundary. in other words, your boundary of x changes as your y changes. if you pick both $0\to r$ as your boundary, it includes (r,r), which is outside your circle. $\endgroup$ – Ching-Ting Wu Nov 26 '16 at 11:07
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As pointed out in the comment, the mistake lies in your thinking that integrating $\sqrt{x^2+y^2}$ over the surface gives you the area, which is wrong. Your integral gives one-fourth of volume of the cone $x^2+y^2=z^2$ in the 3d euclidean octant with height $r\sqrt{2}$. The right integral is $4\int_0^r\int_0^{\sqrt{r^2-y^2}} dx dy$

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If you want to use polar coordinate, consider$$\int_{0}^{2\pi}\int_{0}^{r} r drd\theta$$ which is also the $$\int_{0}^{2\pi}\int_{0}^{r} \sqrt{x^2+y^2} drd\theta$$ you wanted to integrate. Just the boundaries wrong.

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