Double Integral Application with Disks

Question

I have to find the average value of a function $f(x, y) = x + y + x^2 + y^2$

over the disk $0 \le x^2 + y^2 \le 4$, and I am certain this is a double integral problem but I am unsure of the limits of integration for each integrand.

I know the definition of the average value of $f(x,y)$ over a region is $$ f_{avg}=\frac{\iint_Df(x,y)dA}{\iint_D1dA} $$

I am certain I can solve this if I only knew what the limits of integration were. Given the disk is marked as $0 \le x^2 + y^2 \le 4$, my initial presumptions were that the integrand could be written as;

$$ f_{avg} = \frac{ \int_0^{x^2}\int_{y^2}^4 (x+y+x^2+y^2) dydx } { \int_0^{x^2}\int_{y^2}^4 1 dydx } $$

but this is not the case?

Answer 1

First of all, the problem is easier if you switch to polar coordinates.

However, to find the bounds in rectangular coordinates, you do this:

• Given $x^2+y^2\leq4$, what's the smallest and biggest possible value of $y$?

Since $x^2+y^2=\leq4$ is a disk of radius 2, the answer is

• $y$ ranges from $-2$ to $2$.

So, your outer integral is $$\int_{-2}^2\left(\mathrm{inner\ integral}\right) dy$$

Then, you ask this question:

• Given $x^2+y^2\leq 4$, and given a value of $y$, what's the smallest and biggest possible value of $x$?

The answer is:

• $x$ can go as low as $-\sqrt{4-y^2}$, and as high as $+\sqrt{4-y^2}$.

Now you can write down the inner integral: it's

$$\int_{-\sqrt{4-y^2}}^\sqrt{4-y^2}\left(\mathrm{integrand}\right)dx$$

So the whole thing is

$$\int_{-2}^2\int_{-\sqrt{4-y^2}}^\sqrt{4-y^2}\left(\mathrm{integrand}\right)dxdy$$

Answer 2

We can use polar coordinates to make this problem much easier. The disk $D:=\{(x,y):x^2+y^2\leq 4\}$ is defined by a polar angle ranging from $0$ to $2\pi$ and a radius ranging from $0$ to $2$. So we use the polar coordinate transformation to obtain $$\iint_D f(x,y)dA=\int_0^{2\pi}\int_0^2f(rcos(\theta),r\sin(\theta))rdrd\theta$$ where the factor of $r$ is due to the Jacobian of this coordinate transformation. Notice that this makes the computation $\iint_D dA$ quite easy.

We can also do it using Cartesian coordinates, as you were trying to do. The limits of integration in this case are $y$ ranging from the bottom of the disk $D$ (i.e. $-\sqrt{4-x^2}$) to the top of the disk (i.e. $\sqrt{4-x^2}$), and then $x$ ranges from the left end of the disk (i.e. $-2$) to the right end (i.e. $2$). Thus, the integral is: $$\iint_Df(x,y)dA=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}f(x,y)dydx$$