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

Mar 19, 2017

$\int {\int}_{D} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2} {y}^{2}} \mathrm{dA} = 0$

#### Explanation:

We want to evaluate;

$\int {\int}_{D} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2} {y}^{2}} \mathrm{dA}$ where:

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

Which represents the circumference a circle centre $\left(0 , 0\right)$ 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 $\theta = p 2 i$
a ray of fixed length $r = 2$.

And as we convert to Polar coordinates we get:

$x \setminus \setminus \setminus = r \cos \theta = 2 \cos \theta$
$y \setminus \setminus \setminus = r \sin \theta = 2 \sin \theta$
$\mathrm{dA} = \mathrm{dy} \mathrm{dx} \setminus \setminus = r \mathrm{dr} d \theta = 2 \mathrm{dr} d \theta$

So then the integral becomes:

$\int {\int}_{D} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2} {y}^{2}} \mathrm{dA}$
$\text{ } = {\int}_{0}^{2 \pi} {\int}_{2}^{2} {\left(4 - {\left(2 \cos \theta\right)}^{2}\right)}^{- \frac{1}{2} {\left(2 \sin \theta\right)}^{2}} 2 \mathrm{dr} d \theta$

If we look at the inner integral:

${\int}_{2}^{2} {\left(4 - {\left(2 \cos \theta\right)}^{2}\right)}^{- \frac{1}{2} {\left(2 \sin \theta\right)}^{2}} 2 \mathrm{dr}$

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

Hence:

$\int {\int}_{D} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2} {y}^{2}} \mathrm{dA} = {\int}_{0}^{2 \pi} 0 \setminus d \theta$

Which is also $0$