Question

Find `int int_D (4-x^2)^(-1/2y^2) dA` where`D={(x,y) in RR | (x^2+y^2=4 }`?

Answer 1

`int int_D (4-x^2)^(-1/2y^2) dA = 0`

Explanation:

We want to evaluate;

`int int_D (4-x^2)^(-1/2y^2) dA` where:

`D={(x,y) in RR | (x^2+y^2=4 }`

Which represents the circumference a circle centre `(0,0)` and radius `2`. It should be clear that as we are integrating over a region with no area then the result will be `0` (after all a double integral represents the volume over a region, and the region is empty)

We can easily demonstrate this as follows:

If we convert to Polar Coordinates then the region `D` is:

an angle from `theta=0` to `theta=p2i`
a ray of fixed length `r=2`.

And as we convert to Polar coordinates we get:

`x \ \ \ = rcos theta = 2cos theta`
`y \ \ \ = rsin theta = 2 sin theta`
`dA = dy dx \ \ = r dr d theta = 2 dr d theta`

So then the integral becomes:

`int int_D (4-x^2)^(-1/2y^2) dA`
`" " = int_0^(2pi) int_2^2 (4-(2cos theta)^2)^(-1/2(2 sin theta)^2) 2 dr d theta`

If we look at the inner integral:

`int_2^2 (4-(2cos theta)^2)^(-1/2(2 sin theta)^2) 2 dr`

As the upper and lower bounds of integration are the same the integral is zero.

Hence:

`int int_D (4-x^2)^(-1/2y^2) dA = int_0^(2pi) 0 \ d theta`

Which is also `0`