Question

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

Answer 1

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

Explanation:

We want to find:

`I = int int_R \ e^(x^2+y^2) \ dA`

Where `R` is this region, also bounded by the `x`-axis:
graph{ sqrt(9-x^2) [-10, 10, -5, 5] }

If we examine the region `R` we have;

`y = sqrt(9-x^2) => y^2 = 9-x^2 => x^2+y^2 = 3^2`

which is a circle centre `O`, radius `3` (we are only interested in the +ve semi-circle)

We can readily (and logically) perform this integration by converting to Polar coordinates, which leads to the following transformation. The region `R` is:

an angle from `theta=0` to `theta=pi`
a ray from `r=0` to `r=3`

And as we convert to Polar coordinates we get:

`\ \ \ x = rcos theta`
`\ \ \ y = rsin theta`
`dA = dx \ dy = r \ dr \ d theta`

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

`I = int_0^pi int_0^3 \ e^((rcos theta)^2 + (rsin theta)^2) \ r \ dr \ d theta`
`\ \ = int_0^pi int_0^3 \ e^(r^2cos^2 theta + r^2sin^2 theta) \ r \ dr \ d theta`
`\ \ = int_0^pi int_0^3 \ e^(r^2(cos^2 theta + sin^2 theta)) \ r \ dr \ d theta`
`\ \ = int_0^pi int_0^3 \ e^(r^2) \ r \ dr \ d theta`
`\ \ = int_0^pi [1/2 \ e^(r^2)]_0^3 \ d theta`
`\ \ = 1/2 \ int_0^pi (e^9-e^0) \ d theta`
`\ \ = 1/2(e^9-1) \ int_0^pi \ d theta`
`\ \ = 1/2(e^9-1) \ [theta]_0^pi`
`\ \ = 1/2(e^9-1) \ (pi-0)`
`\ \ = (pi(e^9-1))/2`