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Problem: compute area between $y=x$, $x=2$ and $x$ axis, but using Polar coordinates.(without double integral). My idea is to first calculate "normal" type of integral. $$\int_0^2x dx=2$$ Secondly I know that boundaries of new integral are $\pi/4$ (because of $y=x$) and 0 (because of $x$ axis). Also I know that if I want to switch to polar coordinates than $x=r\cos(\alpha)$. But I don't know how to end this problem.

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Consider the polar equation of the vertical leg $\rho(\theta)=\frac{2}{\cos(\theta)}$ and use the formula: $$\frac{1}{2}\int_{\theta=0}^{\pi/4}\rho^2(\theta) \ d\theta=\frac{1}{2}\int_{\theta=0}^{\pi/4}\frac{4}{\cos^2(\theta)}(\theta) \ d\theta= 2\left[\tan(\theta)\right]_0^{\pi/4}=2.$$ Quite a complicated way to compute the area of a right triangle!

P.S. See this webpage for more details about hoew to calculate areas with polar coordinates.

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  • $\begingroup$ Can you explain why 2/cos(x)? $\endgroup$ – josf Aug 8 '16 at 12:22
  • $\begingroup$ @Lovro Sindičić The vertical leg is along the line $x=2$ then use $x=\rho\cos(\theta)$ and find $\rho$. $\endgroup$ – Robert Z Aug 8 '16 at 12:24
  • $\begingroup$ Can you please explain in other way I don't understand. $\endgroup$ – josf Aug 8 '16 at 14:26
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    $\begingroup$ @Lovro Sindičić Look at the first picture here: tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx You have $\alpha=0$ (because of the $x$-axis), $\beta=\pi/4$ (because of the $y=x$) and $\rho=f(\theta)=2/cos(\theta)$ (because of $x=2$, that is $\rho\cos(\theta)=2$ or $\rho=2/\cos(\theta)$. $\endgroup$ – Robert Z Aug 8 '16 at 15:46

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