0
$\begingroup$

I want to integrate a function over a region R, which is determined by the following conditions. $$x^2+y^2\leq1$$ and $$x+y\geq1$$

My region is determined by these two inequalities. I know that the first expression produces a unit circle, which contains all points within the unit circle, including the boundaries.The second expression is a line passing through $y=1$ and with a slope of $-1$.

The region is given to be between the unit circle and the line .I do know what the region looks like , but I do not know how to express it in terms of integration limits, so I can integrate it and find the area, or just do a general integration over this region. I tried to use polar coordinates, but still couldn't figure it out.

Can you guys help me determine the integration limits for this region ?

$\endgroup$
  • $\begingroup$ What function will you be integrating? $\endgroup$ – John Lou Nov 6 '17 at 21:11
  • 3
    $\begingroup$ Do you prefer to use polar coordinates? This can be done with rectangular or polar... $\endgroup$ – G Tony Jacobs Nov 6 '17 at 21:11
  • $\begingroup$ @JohnLou the integrand of my integral is given by dxdy / (x^2+y^2)^2 $\endgroup$ – Viktor Raspberry Nov 6 '17 at 21:12
  • 1
    $\begingroup$ Since the integrand looks like $1/r^4,$ I'd probably try polar coordinates first. Do you know how to write the equation of the line $y=1-x$ in polar coordinates? $\endgroup$ – David K Nov 6 '17 at 21:14
  • $\begingroup$ $x^2 + y^2 \le 1$ is the area inside the circle of unit radius centred. $x + y \ge 1$ is the area above the line $x + y = 1$. Find the intersection between circle and line. $\endgroup$ – A---B Nov 6 '17 at 21:14
3
$\begingroup$

Carrying out this integral in polar coordinates, it is first clear that we need to integrate from $\theta=0$ to $\theta=\frac{\pi}{2}$. The lower limit for $r$ is given by the line, and the upper limit is given by the circle $r=1$. Thus, we need to way to express the line $x+y=1$ in polar coordinates. Using the usual substitution, we can write:

$$r\cos\theta+r\sin\theta=1,$$

or:

$$r=\frac{1}{\cos\theta+\sin\theta}$$

Thus, it looks like we can integrate:

$$\int_0^{\frac{\pi}{2}}\int_{\frac{1}{\cos\theta+\sin\theta}}^1 f(r\cos\theta,r\sin\theta)r\,dr\,d\theta$$

$\endgroup$
1
$\begingroup$

enter image description here

$\int_0^1 \int_{y}^{\sqrt {1-y^2}} f(x,y)\ dx\ dy$ would be one way.

Of course it is symmetric,

$\int_0^1 \int_{x}^{\sqrt {1-x^2}} f(x,y)\ dy\ dx$ would also work.

or you can convert to polar

$x = r\cos\theta\\ y = r\sin \theta\\ dy\ dx = r\ dr\ d\theta$

$x+y \ge1\\ r\cos\theta + r\sin\theta \ge 1\\ r\sqrt 2(\cos\theta - \frac {\pi}{4}) = 1\\ r = \frac {\sqrt 2}{2} \sec (\theta-\frac {\pi}{4})$

$\int_0^{\frac {\pi}{2}} \int_{\sec(\theta - \frac {\pi}{4})}^1 rf(r,\theta)\ dr\ d\theta$

$\endgroup$

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.