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

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

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

Bounded Area =

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

#y# -coordinates are:

# x=-1 => y=1 #

# x=-4 => y=-2 # Giving the coordinates

#(-1,1)# and#(-4,-2)# If in the above diagram we look at an infinitesimally thin horizontal strip (in black) then the limits for

#x# and#y# are:

#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 #

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To find the area bounded by the curves (x = -y^2) and (y = x + 2) using a double integral, you can follow these steps:

- Determine the limits of integration for both (x) and (y) to define the region of interest.
- Set up the double integral over the given region.
- Evaluate the double integral to find the area.

Given (x = -y^2) and (y = x + 2), we need to find the points of intersection to determine the limits of integration.

First, solve (x = -y^2) for (y) to get (y = \sqrt{-x}).

Then, set this expression equal to (y = x + 2) and solve for (x): (\sqrt{-x} = x + 2).

Solve for (x) to find the intersection points.

Now, determine the limits of integration for (x) and (y) based on the intersection points and the shape of the region.

Set up the double integral over the region with the appropriate limits of integration.

Evaluate the double integral to find the area.

Please follow these steps and let me know if you need further assistance.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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