Find the area of the shaded region?

When we first learn to find areas by integration, we take representative rectangles vertically.
The rectangles have base #dx# (a small change in #x#) and heights equal to the greater #y# (the one on upper curve) minus the lesser #y# value (the one on the lower curve). We then integrate from the smallest #x# value to the greatest #x# value.

For this new problem, we could use two such intergrals (See the answer by Jim S), but it is very valuable to learn to turn our thinking #90^@#.

We will take representative rectangles horiontally.
The rectangles have height #dy# (a small change in #y#) and bases equal to the greater #x# (the one on rightmost curve) minus the lesser #x# value (the one on the leftmost curve). We then integrate from the smallest #y# value to the greatest #y# value.

Notice the duality

#{:("vertical ", iff ," horizontal"), (dx, iff, dy), ("upper", iff, "rightmost"), ("lower", iff, "leftmost"), (x, iff, y):}#

The phrase "from the smallest #x# value to the greatest #x# value." indicates that we integrate left to right. (In the direction of increasing #x# values.)

The phrase "from the smallest #y# value to the greatest #y# value." indicates that we integrate bottom to top. (In the direction of increasing #y# values.)

Here is a picture of the region with a small rectangle indicated:

The area is

#int_1^2 (y-1/y^2) dy = 1#

#x=1/y^2#

#y^2=1/x#

#y=sqrtx/x# (we can see from the graph)

#sqrtx/x=x# #<=># #x^2=sqrtx# #<=>#

#x^4-x=0# #<=># #x(x^3-1)=0# #<=># #x=1# (we can also see from the graph)

One of many ways the area of the shaded region can be expressed could be as the area of the triangle #AhatOB=Ω# excluding the cyan area which i will call #color(cyan)(Ω_3)#

Let #Ω_1# be the black area shown in the graph and #color(green)(Ω_2)# the green area shown in the graph.

The area of the small triangle #ChatAD=# #color(green)(Ω_2)# will be:

• #color(green)(Ω_2)=##1/2*1*1=1/2m^2#

#sqrtx/x=2# #<=># #sqrtx=2x# #<=># #x=4x^2#

#<=># #x=1/4#

The area of #Ω_1# will be:

#int_(1/4)^1(2-sqrtx/x)dx=2[x]_(1/4)^1-2[sqrtx]_(1/4)^1=#

#2(1-1/4)-2(1-sqrt(1/4))=6/4-2(1-1/2)#

#=3/2-1=1/2m^2#

As a result, the shaded area will be

• #Ω_1##+color(green)(Ω_2)##=1/2+1/2=1m^2#