Question

Find the average value of the function f(x, y) = 7/ x2 + y2 on the annular region a2 ≤ x2 + y2 ≤ b2, where 0 < a <

b.

Answer 1

To find the average value of the function f(x, y) = 7/ (x^2 + y^2) on the annular region a^2 ≤ x^2 + y^2 ≤ b^2, you need to evaluate the integral ∫∫(7/(x^2 + y^2)) dA over the annular region. Convert the integral to polar coordinates and then evaluate it by integrating with respect to r and θ. Divide the result by the area of the annular region to find the average value of the function.

Step-by-step explanation:

Step 1: Evaluate the integral

To find the average value of the function, we need to evaluate the integral of the function over the annular region. The integral of f(x, y) = 7/(x^2 + y^2) over the annular region is given by:

∫∫(7/(x^2 + y^2)) dA

Step 2: Convert to polar coordinates

In order to evaluate the integral, we can convert to polar coordinates. The annular region can be redefined in terms of polar coordinates as:

a^2 ≤ r^2 ≤ b^2

Step 3: Evaluate the integral using polar coordinates

By substituting x = rcosθ and y = rsinθ in the integral, we obtain:

∫∫(7/(r^2)) r dr dθ

Step 4: Integrate over the annular region

Integrate the inner integral with respect to r, and then integrate the outer integral with respect to θ from 0 to 2π, to cover the entire annular region.

Step 5: Calculate the average value

Finally, divide the result by the area of the annular region to find the average value of the function.

Answer 2

Denote the annular region by
`A`. The average value of
`f(x,y)=\frac7{x^2+y^2}` over
`A` is given by


`(\displaystyle\iint_A f(x,y)\,\mathrm dA)/(\displaystyle\iint\mathrm dA)`

For both integrals, convert the region to polar coordinates. We have


`\displaystyle\iint_Af(x,y)\,\mathrm dA=\int_(\theta=0)^(\theta=2\pi)\int_(r=a)^(r=b)\frac7{r^2}r\,\mathrm dr\,\mathrm d\theta`

`=14\pi\displaystyle\int_(r=a)^(r=b)\frac{\mathrm dr}r`

`=14\pi\ln\frac ba`


`\displaystyle\iint_A\mathrm dA=\int_(\theta=0)^(\theta=2\pi)\int_(r=a)^(r=b)r\,\mathrm dr\,\mathrm d\theta`

`=\pi(b^2-a^2)`

and so the average value is


`(14)/(b^2-a^2)\ln\frac ba`