# How to calculate this area in polar coordinates

I would like to calculate the area of the region between the $y=x^2$ and $y=x$ using double integrals. The problem is I couldn't find the limits of the variable $r$ in this integral, because the line $y=x$ in polar coordinates is $\theta=\pi/4$ which doesn't depend of $r$.

• You have a double integral, one integral on $r$, and one on $\theta$. This $\pi/4$ would go on one of your limits for $\theta$. What is your exact region? The region where $y\leq x$ and $y\geq x^2$?. – John Doe Apr 27 '17 at 0:10
• @JohnDoe I didn't understand, could you give me the explicit limits of this integral? Thank you – user42912 Apr 27 '17 at 0:12

Assuming that the area you're after is the region enclosed by these two curves: Then $\theta=\pi/4$ on the upper limit, and on the lower limit it is $0$, since this is the lowest you can go in this quadrant.

Meanwhile for the $r$ limits, given a fixed $\theta$, how far can you go outwards from $0$ until you hit the curve $y=x^2$? This can be solved as follows:

$$y=x^2\implies r\sin\theta=r^2\cos^2\theta\implies r=\tan\theta\sec\theta$$

So $r$ goes up to here. So we have $$\int_0^{\pi/4} \left(\int_0^{\tan\theta\sec\theta}rdr\right)d\theta$$

• Thank you very much, great answer! – user42912 Apr 27 '17 at 14:23
• Where did you plot this graph? – user42912 Apr 27 '17 at 14:35
• @user42912 On MATLAB – John Doe Apr 27 '17 at 16:05
• I knew Wolfram Alpha and Mathematica as well. I was just curious because your plot seems different from other softwares I know. – user42912 Apr 27 '17 at 16:31
• Oh, yes I find MATLAB quite easy for producing graphs – John Doe Apr 27 '17 at 16:32