Question

Find the area of the shaded region?

`x=y` and `x=1/y^2`

Answer 1

Please see below.

Explanation:

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`

Answer 2

Area of the shaded region is `1m^2`

Explanation:

`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`