What is the area bounded by the curves? : # 4x + y^2 = 32 # and # x=y #

We have:

# 4x + y^2 = 32 #
# x=y #

The graphs are as follows:

To find the coordinates of intersection:

# 4x + x^2 = 32 #
# :. (x-4)(x+8) = 0 #
# => x=4, -8#

So the intersection coordinates are:

# (-8,-8)# and #(4,4)#

We can calculate the bounded area (shaded) by integrating either wrt #x# or wrt #y#, the latter being easier.

Method 1: Integrating wrt #y#

if we integrate with infinestimall thin horiozontal striops then we find the strips are bounded by #y=x# on the left and by #4x + y^2 = 32# on the right, and we integrate from the lower point of intersection# (-8,-8)# to the upper point of intersection #(4,4)#.

Thus, the area is given by:

# A = int_alpha^beta \ f(y) \ dy #
# \ \ \ = int_(y=-8)^(y=4) \ ( (32-y^2)/4 ) -(y) \ dy #
# \ \ \ = int_(-8)^(4) \ 8 - y^2/4 -y \ dy #
# \ \ \ = [ 8y - y^3/12 -y^2/2 ]_(-8)^(4) #
# \ \ \ = (32-16/3-8) - (-64+128/3-32) #
# \ \ \ = 56/3 - (-160/3) #
# \ \ \ = 72 #

Method 2: Integrating wrt #x#

if we integrate with infinestimall thin vertical striops then we find the strips are bounded by #4x + y^2 = 32# at the bottom and #y=x# at the top but only when #x# is between the lower point of intersection# (-8,-8)# to the upper point of intersection #(4,4)#

We also need to include a portion that is bounded by #4x + y^2 = 32# at the top and bottom between the upper point of intersection and the vertex (at #x=8#).

Thus, the area is split into two seperate integrals:

# A = A_1 + A_2 #

Where:

# A_1 = int_(x=-8)^(x=4) (x)-(-sqrt(32-4x)) \ dx #
# A_2 =int_(x=4)^(x=8) \ (sqrt(32-4x) - (-sqrt(32-4x) ) \ dx #

First we calculate #A_1#:

# A_1 = int_(-8)^(4) x+sqrt(32-4x) \ dx #
# \ \ \ \ = [x^2/2 -(4(8-x)^(3/2))/3 ]_(-8)^(4) #
# \ \ \ \ = (8-32/3) - (32 - 256/3) #
# \ \ \ \ = (-8/3) - (-160/3) #
# \ \ \ \ = 152/3 #

Then, #A_2#

# A_2 =int_(4)^(8) \ 2sqrt(32-4x) \ dx #
# \ \ \ \ = 2 int_(4)^(8) \ sqrt(32-4x) \ dx #
# \ \ \ \ = 2 [-(4(8-x)^(3/2))/3 ]_(4)^(8) #
# \ \ \ \ = 2 { (0) - (-32/3) } #
# \ \ \ \ = 64/3 #

So, the total area is:

# A = A_1 + A_2 #
# \ \ \ = 152/3 + 64/3 #
# \ \ \ = 72 #