# Find the area bounded by # x = -y^2 # and # y = x+2# using a double integral?

##
(portions of this question have been edited or deleted!)

(portions of this question have been edited or deleted!)

##### 1 Answer

Bounded Area =

#### Explanation:

Based on the sketch. we are looking for a double integral solution to calculate the area bounded by the curves:

# x = -y^2 #

# y = x+2 = > x=y-2#

The points of intersection are the solution of the equation:

# x = -(x+2)^2 #

# :. x = -(x^2+4x+4) #

# :. x^2+5x+4 = 0 # :. (x+1)(x+4) = 0# # :. x=-1, -4#

The corresponding

# x=-1 => y=1 #

# x=-4 => y=-2 #

Giving the coordinates

If in the above diagram we look at an infinitesimally thin horizontal strip (in black) then the limits for

#x# varies from#y-2# to#-y^2#

#y# varies from#-2# to#1#

And so we can represent the bounded are by the following double integral:

# A = int int_R dA #

# \ \ \ = int_-2^1 \ int_(y-2)^(-y^2) \ dx \ dy #

We can calculate the inner integral:

# int_(y-2)^(-y^2) \ dx = [x]_(y-2)^(-y^2) #

# " " = (-y^2) - (y-2) #

# " " = -y^2 - y+2 #

# " " = -(y^2 + y-2) #

And so:

# A = int_-2^1 \ int_(y-2)^(-y^2) \ dx \ dy #

# \ \ \ = int_-2^1 -(y^2 + y-2) \ dy #

# \ \ \ = - int_-2^1 y^2 + y-2 \ dy #

# \ \ \ = - [ y^3/3 + y^2/2-2y ]_-2^1 #

# \ \ \ = - {(1/3+1/2-2)-(-8/3+2+4)} #

# \ \ \ = - {(-7/6)-(10/3)} #

# \ \ \ = 9/2 #