Chapter 8.2, Problem 11E

To determine

To find the area enclosed by the graph of the two curves.

Answer to Problem 11E

The area enclosed by the graphs of the two curve is 7.542 squared units.

Explanation of Solution

Concept used: Area between curves.

If f and g are continuous with $f\left(x\right)\ge g\left(x\right)$ throughout [a,b], then the area between

The curves $y=f\left(x\right)\text{ and }y=g\left(x\right)$ from a to b is the integral of $\left[f-g\right]$ from ato b.

  $\begin{array}{l}A={\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]}dx\\ \end{array}$

Given information:

The given function is, $y^2=x+1\text{ , }{y}^2=3-x.$

Calculation: The area enclosed by the graphs of the two curves is obtained as,

Set up each equation in term of y,

  $\begin{array}{l}{y}^2=x+1y^2=3-x\\ x=y^2-1x=3-y^2\end{array}$

Draw the function into the graph to see which of the functions is on top and which one is on bottom.

  

So, the blue curve is $x=3-y^2$ which one is on top, while the red curve is $x=y^2-1$

Is on bottom.

Set both the equations equal to each other,

  $\begin{array}{l}{y}^2-1=3-y^2\left(\text{solve for the limits}\right)\\ 2y^2=4\\ y^2=2\\ y=\pm \sqrt{2}\end{array}$

The curves $y^2=x+1$ and $y^2=3-x$ intersect at $y=\pm \sqrt{2}.$

Now, area bounded by these region is,

  $\begin{array}{l}\text{A}={\displaystyle {\int }_a^b\left[f\left(t\right)-g\left(t\right)\right]}dt\\ ={\displaystyle {\int }_{-\sqrt{2}}^{\sqrt{2}}\left[\left(3-y^2\right)-\left(y^2-1\right)\right]}dy\\ ={\displaystyle {\int }_{-\sqrt{2}}^{\sqrt{2}}\left[4-2y^2\right]}dy\\ ={\left[4y-\frac{2y^3}{3}\right]}_{-\sqrt{2}}^{\sqrt{2}}\\ =\left[4\left(\sqrt{2}\right)-\frac{2{\left(\sqrt{2}\right)}^3}{3}-\left(-4\sqrt{2}-\frac{2{\left(-\sqrt{2}\right)}^3}{3}\right)\right]\\ =4\sqrt{2}+4\sqrt{2}-\frac{4\sqrt{2}}{3}-\frac{4\sqrt{2}}{3}\\ =8\sqrt{2}-\frac{8\sqrt{2}}{3}\\ \approx \text{7}\text{.542}\end{array}$

Hence, the area enclosed by the graphs of the two curve is 7.542 squared units.