Chapter 8.2, Problem 40E

a.

To determine

To Find:

To find the area region between the curves $y=3-x^2$ and the line $y=-1$ with respect to x .

Answer to Problem 40E

The area of the region bounded by curve and line is $\frac{32}{3}$ .

Explanation of Solution

Given information:

The given function is $y=3-x^2$ and $y=-1$

The red curve is the graph of $y=3-x^2$ and the blue line is the graph of $y=-1$ .

  

Therefore, the area of the region bounded by the curve and line is,

  $\begin{array}{l}A={\displaystyle \underset{-2}{\overset{2}{\int }}\left(3-x^2+1\right)dx}\\ A={\left[3x-\frac{x^3}{3}+x\right]}_{-2}^2\\ A=\left[3\left(2\right)-\frac{2^3}{3}+2\right]-\left[3\left(-2\right)-\frac{{\left(-2\right)}^3}{3}-2\right]\\ A=\left[6-\frac{8}{3}+2\right]-\left[-6+\frac{8}{3}-2\right]\\ A=6-\frac{8}{3}+2+6-\frac{8}{3}+2\\ A=16-\frac{16}{3}\\ A=\frac{32}{3}\end{array}$

Therefore, the area of the region bounded by the curve and line is $\frac{32}{3}$ .

b.

To determine

To Find:

To find the area region between the curves $y=3-x^2$ and the line $y=-1$ with respect to y .

Answer to Problem 40E

The area of the region bounded by curve and line is $\frac{32}{3}$ .

Explanation of Solution

Given information:

The given function is $y=3-x^2$ and $y=-1$ .

Formula used:

  ${\displaystyle \int x^a}dx=\frac{x^{a+1}}{a+1}$

The red curve is the graph of $y=3-x^2$ and the blue line is the graph of $y=-1$ .

  

Therefore, the area of the region bounded by the curve and line is,

  $A=2{\displaystyle \underset{-1}{\overset{3}{\int }}\sqrt{3-y}dy}$

Substitute $u=3-y$ ,then the interval runs from

  $\begin{array}{l}y=-1\Rightarrow u=4\\ y=3\Rightarrow u=0\end{array}$

Hence the integral becomes,

  $\begin{array}{l}=2{\displaystyle {\int }_4^0-}\sqrt{u}du\\ =-2{\displaystyle {\int }_0^4-}\sqrt{u}du\\ =2{\displaystyle {\int }_0^4\sqrt{u}du}\end{array}$

Now using the formula, ${\displaystyle \int x^a}dx=\frac{x^{a+1}}{a+1}$ ,

  $\begin{array}{l}=2\left({\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^4\right)\\ =2{\left[\frac{2}{3}{u}^{\frac{3}{2}}\right]}_0^4\\ =2\left\{\left[\frac{2}{3}{\left(4\right)}^{\frac{3}{2}}\right]-\left[\frac{2}{3}{\left(0\right)}^{\frac{3}{2}}\right]\right\}\\ =2\left(\frac{16}{3}\right)\\ =\frac{32}{3}\end{array}$

Therefore, the area of the region bounded by the curve and line is $\frac{32}{3}$ .