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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.