Hello, mesterpa!

It seems to be easy enough . . .

Can someone help me with this double integral please.

∫∫ e^{-r²} r dr dθ

. . whereris bounded by 0 and a

. . and θ is bounded by 0 and π/2

The inner integral is: .∫ e^{-r²} r dr

Let: u = -r² . → . du = -2r dr

And we have: .-½∫ e^u du .= .-½ e^u . → . -½e^{-r²}

Now evaluate for r = 0 to r = a.

Then integrate with respect to θ and evaluate.