# Find # int int \ e^(x^2+y^2) \ dA # over the region bounded by #y = sqrt(9-x^2)#?

# (pi(e^9-1))/2 #

We want to find:

And as we convert to Polar coordinates we get:

And so we get (by performing the inner integration first, followed by the outer):

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To find ( \iint e^{x^2+y^2} , dA ) over the region bounded by ( y = \sqrt{9-x^2} ), we first need to set up the integral using polar coordinates.

The region can be described in polar coordinates as (0 \leq r \leq 3) and (0 \leq \theta \leq \pi) (half of the circle (x^2 + y^2 = 9) in the positive y-axis).

So, the integral becomes:

[ \iint e^{x^2+y^2} , dA = \int_{0}^{\pi} \int_{0}^{3} e^{r^2} \cdot r , dr , d\theta ]

Now, solve this integral.

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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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