Chapter 7, Problem 49RE

a.

To determine

To find:The equation of the parabola with $y$ -intercepts.

Answer to Problem 49RE

The equation of the parabola with $y$ -intercepts is $y=-\frac{5}{4}{x}^2+5$ .

Explanation of Solution

Given information:

  $y$ -intercept = $5$

  $x$ -intercept = $\pm 2$

Calculation:

Let's plot the intercept and connect them first in order to create the model for the usual slice, as shown below.

  

The standard equation of the parabola is given by,

  $y=a{x}^2+bx+c$

Forming the equations by using the given intercepts,

At $\left(-2,0\right)$ ,

  $\begin{array}{l}y=a{x}^2+bx+c\\ 0=a{\left(-2\right)}^2+b\left(-2\right)+5\\ 0=4a-2b+5\end{array}$

At $\left(2,0\right)$ ,

  $\begin{array}{l}y=a{x}^2+bx+c\\ 0=a{\left(2\right)}^2+b\left(2\right)+5\\ 0=4a+2b+5\end{array}$

Now solving the above two equations to get,

  $\begin{array}{l}\underset{¯}{\begin{array}{l}0=4a-2b+5\\ 0=4a+2b+5\end{array}}\\ 0=0a-4b-0\\ b=0\end{array}$

Substituting the value of $b$ to get the value of $a$ ,

  $\begin{array}{c}0=a{\left(-2\right)}^2+b\left(-2\right)+5\\ =4a-2\left(0\right)+5\\ =4a+5\\ a=-\frac{5}{4}\end{array}$

Substituting the value of $a\text{ and }b$ in the standard equation to get,

  $\begin{array}{c}y=-\frac{5}{4}{x}^2+0x+5\\ =-\frac{5}{4}{x}^2+5\end{array}$

Therefore, the equation of the parabolais $y=-\frac{5}{4}{x}^2+5$ .

b.

To determine

To find: The volume of the Bundt cake.

Answer to Problem 49RE

The volume of the Bundt cakeis $V=335.1032$ .

Explanation of Solution

Given information:

  $y$ -intercept = $5$

  $x$ -intercept = $\pm 2$

Calculation:

The line is get revolved at about $x=4$ .

From this, let’s know

Radius of the shell = $4-x$

Height = $5-\frac{5}{4}{x}^2$

By using the cylindrical shell method,

  $\begin{array}{c}V={\displaystyle {\int }_{-2}^{2}2\pi \left(4-x\right)\left(5-\frac{5}{4}{x}^2\right)dx}\\ =2\pi {\displaystyle {\int }_{-2}^2\left(20-5x^2-5x+\frac{5x^3}{4}\right)dx}\\ =2\pi {\displaystyle {\int }_{-2}^2\left(20\right)dx}-2\pi {\displaystyle {\int }_{-2}^2\left(5x^2\right)dx}-2\pi {\displaystyle {\int }_{-2}^2\left(5x\right)dx}+2\pi {\displaystyle {\int }_{-2}^2\left(\frac{5x^3}{4}\right)dx}\\ =V_1-V_2-V_3+V_4\end{array}$

To find: $V_1$

  $\begin{array}{c}{V}_1=2\pi {\displaystyle {\int }_{-2}^2\left(20\right)dx}\\ =2\pi {\left.\left(20x\right)\right|}_{-2}^2\\ =40\pi \left(2-\left(-2\right)\right)\\ =40\pi \left(4\right)\\ =160\pi \end{array}$

To find: $V_2$

  $\begin{array}{c}{V}_2=2\pi {\displaystyle {\int }_{-2}^2\left(5x^2\right)dx}\\ =10\pi {\left.\left(\frac{x^3}{3}\right)\right|}_{-2}^2\\ =10\pi \left(\frac{2^3}{3}-\left(\frac{-2^3}{3}\right)\right)\\ =10\pi \left(\frac{16}{3}\right)\\ =\frac{160}{3}\pi \end{array}$

To find: $V_3$

  $\begin{array}{c}{V}_3=2\pi {\displaystyle {\int }_{-2}^2\left(5x\right)dx}\\ =10\pi {\left.\left(\frac{x^2}{2}\right)\right|}_{-2}^2\\ =10\pi \left(\frac{2^2}{2}-\left(\frac{-2^2}{2}\right)\right)\\ =10\pi \left(0\right)\\ =0\end{array}$

To find: $V_4$

  $\begin{array}{c}{V}_4=2\pi {\displaystyle {\int }_{-2}^2\left(\frac{5x^3}{4}\right)dx}\\ =2\pi {\left.\left(\frac{5x^4}{4}\right)\right|}_{-2}^2\\ =2\pi \left(\frac{5{\left(2\right)}^4}{4}-\left(\frac{5{\left(2\right)}^4}{4}\right)\right)\\ =2\pi \left(5-5\right)\\ =0\end{array}$

Substituting the above values in the equation to get,

  $\begin{array}{c}V=V_1-V_2-V_3+V_4\\ =160\pi -\frac{160}{3}\pi -0+0\\ =\frac{320}{3}\pi \\ V=335.1032\end{array}$

Therefore, the total volume is $V=335.1032$ .