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

##### 1 Answer

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

#### Explanation:

We want to find:

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

Where

graph{ sqrt(9-x^2) [-10, 10, -5, 5] }

If we examine the region

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

which is a circle centre

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

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 #