# Find the area of the shaded region?

## $x = y$ and $x = \frac{1}{y} ^ 2$

May 20, 2018

#### Explanation:

When we first learn to find areas by integration, we take representative rectangles vertically.
The rectangles have base $\mathrm{dx}$ (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 ${90}^{\circ}$.

We will take representative rectangles horiontally.
The rectangles have height $\mathrm{dy}$ (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

${\int}_{1}^{2} \left(y - \frac{1}{y} ^ 2\right) \mathrm{dy} = 1$

May 20, 2018

Area of the shaded region is $1 {m}^{2}$

#### Explanation:

$x = \frac{1}{y} ^ 2$

${y}^{2} = \frac{1}{x}$

$y = \frac{\sqrt{x}}{x}$ (we can see from the graph)

$\frac{\sqrt{x}}{x} = x$ $\iff$ ${x}^{2} = \sqrt{x}$ $\iff$

${x}^{4} - x = 0$ $\iff$ $x \left({x}^{3} - 1\right) = 0$ $\iff$ $x = 1$ (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 $C \hat{A} D =$ color(green)(Ω_2) will be:

• color(green)(Ω_2)=$\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} {m}^{2}$

$\frac{\sqrt{x}}{x} = 2$ $\iff$ $\sqrt{x} = 2 x$ $\iff$ $x = 4 {x}^{2}$

$\iff$ $x = \frac{1}{4}$

The area of Ω_1 will be:

${\int}_{\frac{1}{4}}^{1} \left(2 - \frac{\sqrt{x}}{x}\right) \mathrm{dx} = 2 {\left[x\right]}_{\frac{1}{4}}^{1} - 2 {\left[\sqrt{x}\right]}_{\frac{1}{4}}^{1} =$

$2 \left(1 - \frac{1}{4}\right) - 2 \left(1 - \sqrt{\frac{1}{4}}\right) = \frac{6}{4} - 2 \left(1 - \frac{1}{2}\right)$

$= \frac{3}{2} - 1 = \frac{1}{2} {m}^{2}$

As a result, the shaded area will be

• Ω_1+color(green)(Ω_2)$= \frac{1}{2} + \frac{1}{2} = 1 {m}^{2}$