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

Question

Emma Aldridge · Accepted Answer

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

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#

Axel Alvarez · Answer

undefined The integral $ \iint_D (4-x^2)^{-\frac{1}{2}y^2} \, dA $ over the region $ D = \{(x,y) \in \mathbb{R}^2 \, | \, x^2+y^2=4 \} $ is equal to $ \frac{\pi}{2} $.