Chapter 8.2, Problem 41E

a.

To determine

To graph:

To sketch the region and draw a line $y=c$ across the parabola $y=x^2$ and the line $y=4$ .

Explanation of Solution

Given information:

The given function is $y=x^2$ and $y=4$ .

The red curve is the graph of $y=x^2$ and the blue line is the graph of $y=4$

  

Interpretation:

The green line is the graph of $y=c$ . The line $y=c$ intersects the parabola at $\left(1,1\right)$ and the line at $\left(4,4\right)$ .

b.

To determine

To Find:

To find the value of c by integrating with respect to y.

Answer to Problem 41E

The value of c is $c\approx 2.5198421$ .

Explanation of Solution

Given information:

The given function is $y=x^2$ and $y=4$ .

The red curve is the graph of $y=x^2$ and the blue line is the graph of $y=4$ and the green line is the graph of $y=c$ .

  

The co-ordinates of the points where the line and parabola intersect is

  $\begin{array}{l}c=x^2\\ x=\pm \sqrt{c}\end{array}$

Therefore, the co-ordinates of the points is $\left(-\sqrt{c},c\right)$ and $\left(\sqrt{c},c\right)$

Hence, the value of c can be determined by,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{c}{\int }}{y}^{\frac{1}{2}}}dy={\displaystyle \underset{c}{\overset{4}{\int }}{y}^{\frac{1}{2}}dy}\\ {\left[\frac{2}{3}{y}^{\frac{3}{2}}\right]}_0^c={\left[\frac{2}{3}{y}^{\frac{3}{2}}\right]}_c^4\\ \frac{2}{3}{\left(c\right)}^{\frac{3}{2}}=\left(\frac{2}{3}{\left(4\right)}^{\frac{3}{2}}\right)-\left(\frac{2}{3}{\left(c\right)}^{\frac{3}{2}}\right)\\ \frac{2}{3}{c}^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}{c}^{\frac{3}{2}}\\ 2\left(\frac{2}{3}{c}^{\frac{3}{2}}\right)=\frac{16}{3}\\ \left(\frac{4}{3}{c}^{\frac{3}{2}}\right)=\frac{16}{3}\\ c^{\frac{3}{2}}=4\\ c={\left(4\right)}^{\frac{2}{3}}\\ c=2.51984\end{array}$

Therefore, the value of c is $c\approx 2.5198421$ .

c.

To determine

To Find:

To find the value of c by integrating with respect to x.

Answer to Problem 41E

The value of c is $c\approx 2.5198421$ .

Explanation of Solution

Given information:

The given function is $y=x^2$ and $y=4$ .

The red curve is the graph of $y=x^2$ and the blue line is the graph of $y=4$ and the green line is the graph of $y=c$ .

  

The interval runs from $-\sqrt{c}$ to $\sqrt{c}$ and the region bounded above by $y=c$ and below $y=x^2$ .

We need to find the value of c so that the area below the line $y=c$ is

  $A=\frac{1}{2}\cdot \frac{32}{3}=\frac{16}{3}$

Therefore, the value of c can be determined by,

  $\begin{array}{l}{\displaystyle \underset{-\sqrt{c}}{\overset{\sqrt{c}}{\int }}c-x^2}dx=\frac{16}{3}\\ {\left[cx-\frac{x^3}{3}\right]}_{-\sqrt{c}}^{\sqrt{c}}=\frac{16}{3}\\ \left[c\sqrt{c}-\frac{{\left(\sqrt{c}\right)}^3}{3}\right]-\left[c\left(-\sqrt{c}\right)-\frac{{\left(-\sqrt{c}\right)}^3}{3}\right]=\frac{16}{3}\\ c\sqrt{c}-\frac{c^{\frac{3}{2}}}{3}+c\sqrt{c}-\frac{c^{\frac{3}{2}}}{3}=\frac{16}{3}\\ 2c\sqrt{c}-2\frac{c^{\frac{3}{2}}}{3}=\frac{16}{3}\\ 2c^{\frac{3}{2}}-2\frac{c^{\frac{3}{2}}}{3}=\frac{16}{3}\\ \frac{4}{3}{c}^{\frac{3}{2}}=\frac{16}{3}\\ c^{{}^{\frac{3}{2}}}=4\\ c={\left(4\right)}^{\frac{2}{3}}\\ c=2.51984\end{array}$

Therefore, the value of c is $c\approx 2.5198421$ .