Find the area of the shaded region?

x=y and x=1/y^2

enter image source here

2 Answers
May 20, 2018

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:

enter image source here

The area is

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

May 20, 2018

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)

enter image source here

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