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

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#