Chapter 7, Problem 8RE

To determine

To graph: The area can be calculated using a graph $y={\left(1-\sqrt{x}\right)}^2$ .

Explanation of Solution

Given Information: To determine the size of the area that is bounded by the $y=0,x=0$ and the curve $\sqrt{x}+\sqrt{y=1}$ . Since this curve is supplied in implicit form, let's first rewrite it in explicit form so that the area can be calculated. Therefore,

  $y={\left(1-\sqrt{x}\right)}^2$

These graphs' sketches make it easier to identify specific issues.

So let's draw the graphs into Geiger and highlight the necessary area in blue.

  

Interpretation : By carefully examining the provided image, you may deduce that the region is bounded above by the curve $y={\left(1-\sqrt{x}\right)}^2$ and below by the line $y=0$ from $x=0$ to $x=1$ . It is simple to verify that the intersection is formed by the points $\left(0,1\right)$ and $\left(1,0\right)$ .

The definition (page 397) that area of the region is represented as sets the integration's boundaries.

  $Area={\displaystyle \underset{0}{\overset{1}{\int }}\left[{\left(1-\sqrt{x}\right)}^2-0\right]}dx$

Simply square the sub integral function to solve the previous integral, then carry on with the calculation:

  $\begin{array}{c}Area={{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-\sqrt{x}\right)}}^{2}dx\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(1-2\sqrt{x+x}\right)}dx\\ =\left[x-2\cdot \frac{2}{3}\cdot x^{3/2}+\frac{x^2}{2}\right]\left|{}_0^1\right.\\ =\left(1-\frac{4}{3}+\frac{1}{2}\right)-0\\ =\frac{1}{6}\end{array}$

The region's area is equal $\frac{1}{6}.$