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

Question

William Abdalla · Accepted Answer

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

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 #

Emma Aldridge · Answer

undefined 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.