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

##### 1 Answer

# 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

We can easily demonstrate this as follows:

If we convert to Polar Coordinates then the region

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