change the order of integration for the following integral from dydx to dxdy, and from dydx to polar coordinates. $$ \int \int f(x,y) dydx$$

where $$ 0≤y≤(-x^2)+2 $$ $$ 0≤x≤1$$

From dydx to dxdy $$ \int \int f(x,y) dxdy + \int \int f(x,y) dxdy$$

First integral
$$ 0≤x≤1 $$ $$ 0≤y≤1 $$

Second integral $$ 0≤x≤\sqrt(2-y) $$ $$ 0≤y≤1 $$

I'm not sure about the $\sqrt(2-y)$ for the bounds of x for the second integral.


I'm having more trouble converting this into polar coordinates though. I think I can leave the first integral as it is in terms of dxdy, because the region is a rectangle. Is there any way to switch this rectangular region into polar coordinates?

For the second integral

$$ 0≤r≤1 $$ $$ 0≤\theta≤\pi/2 $$

$$ \int \int f(x,y) dxdy + \int \int f(r,\theta) rdrd\theta$$


your work fipping the order of integration is correct.

coverting to polar -- it is going to get messy.

$x = r \cos \theta\\ y = r \sin \theta$

Inside the rectangular region.

$\theta \in [0,\frac{\pi}{4})\\ x =1\\ r \cos\theta = 1\\ r = \sec\theta$

Inside the parabola

$\theta \in [\frac{\pi}{4},\frac{\pi}{2}]\\ y = -x^2 + 2\\ r \sin \theta = - r^2 cos^2 \theta + 2\\ r^2 \cos^2 \theta + r \sin \theta - 2 = 0\\ r = \frac {-\sin\theta + \sqrt{sin^2 \theta + 8 cos^2 \theta}}{2 cos^2 \theta}$

I wouldn't want to integrate that, but it would be a limit.

  • $\begingroup$ I just noticed for my rectangular region (second integral) for y I have 0≤y ≤1. Should it be 1≤y ≤2? $\endgroup$ – says Jun 23 '16 at 19:08
  • $\begingroup$ But is 1≤y ≤2 'more correct'? I think they would produce the same result, as the difference between upper and lower bounds is still 1. $\endgroup$ – says Jun 23 '16 at 19:30
  • $\begingroup$ Sorry, confusing myself, I was thinking about $x's.$ you are right. $y$ in $[0,1]$ is the rectangle. $y$ in $[1, 2]$ is the parabola. $\endgroup$ – Doug M Jun 23 '16 at 19:34


Where the overlapping region of red and blue is the desired region.

Cut the region using the green line ($x-y=0$) instead.

The upper region becomes $\displaystyle\int_{\pi/4}^{\pi/2}\int_0^{(\sqrt{1+7\cos^2\theta}-\sin\theta)/(2\cos^2\theta)}f(r,\theta)\ r\ \mathrm dr\ \mathrm d\theta$.

The lower region becomes $\displaystyle\int_{0}^{\pi/4}\int_0^{\sec\theta}f(r,\theta)\ r\ \mathrm dr\ \mathrm d\theta$.

Note: $(\sqrt{1+7\cos^2\theta}-\sin\theta)/(2\cos^2\theta)$ is the positive solution of $(r\cos\theta)^2+(r\sin\theta)=2$

  • $\begingroup$ It looks like you have updated your post. I have something similar. $\endgroup$ – Doug M Jun 23 '16 at 16:41

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