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#

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#
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Answer 2

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} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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