Determining integration limits for the following region.

Question

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 ?

Answer 1

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:

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

Answer 2

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

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

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

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