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

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

We want to evaluate;

We can easily demonstrate this as follows:

And as we convert to Polar coordinates we get:

So then the integral becomes:

If we look at the inner integral:

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

Hence:

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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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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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