Q. 8 - Calculus

Question

For a four-disk approximation of the volume of the solid obtained from the region between $f\left(x\right)=x^2$ and the $y$ axis between $y=0$ and $y=4$by rotating around the $y$-axis, illustrate and calculate

(a) $\Delta y$ and each $y_{ki}$

(b) some $y_k^{*}$ in each subinterval$\left[y_{k-1},y_k\right]$;

(c) each $f^{-1}\left(y_k^{*}\right)$;

(d) the volume of the second disk.

Write each of the limits in Exercises 9-11 in terms of definite integrals, and identify a solid of revolution whose volume is represented by that definite integral.

Step-by-Step Solution

Answer

Part (a) $y_1=1, y_2=2, y_3=3, y_4=4$

Part (b) $y_1^{*}=0.5, y_2^{*}=1.5, y_3^{*}=2.5, y_4^{*}=3.5$

Part (c) $0.71, 1.22, 1.58, 1.87$

Part (d) $1.5\mathrm{\pi }$

Part (a) Step 1: Given information

In between ranges $y=0$ and $y=4$, the function is $f\left(x\right)=x^2$

The goal is to use a midpoint sum with four rectangles to approximate the area of this region using a Riemann sum.

Part (a) Step 2: Calculation

The strategy would include cutting the area into four rectangles. Each rectangle's width is calculated as,

$\begin{array}{rcl}\Delta y& =& \frac{b-a}{4}\\ & =& \frac{4-0}{4}\\ & =& 1\end{array}$

The first rectangle's beginning value is 0. Each rectangle's end point, denoted by $k^{\text{th }}$, can be thought of as $y_{\lambda }$. Use the relation below to ascertain these endpoints.

$y_k=y_{\lambda -1}+\Delta y$

Use the different values of " k ' to determine different values of $y_{\lambda }$.

The first value will be $y_6=0$.

Find the following end value using the value $k=1$.

$\begin{array}{rcl}{y}_1& =& y_0+1\\ & =& 0+1\\ & =& 1\end{array}$

Use the value $k=2$ to determine the next end value.

$\begin{array}{rcl}{y}_2& =& y_1+1\\ & =& 1+1\\ & =& 2\end{array}$

Use the value $k=3$ to determine the next end value.

$\begin{array}{rcl}{y}_3& =& y_2+1\\ & =& 2+1\\ & =& 3\end{array}$

Use the value $k=4$ to determine the next end value.

$\begin{array}{rcl}{y}_4& =& y_3+1\\ & =& 3+1\\ & =& 4\end{array}$

Part (b) Step 1: Given information

In between ranges $y=0$ and $y=4$, the function is $f\left(x\right)=x^2$.

The goal is to use a midpoint sum with four rectangles to approximate the area of this region using a Riemann sum.

Part (b) Step 2: Calculation

The interval of division for each rectangle using this width is given as $[0,1],[1,2],[2,3],[3,4]$.

The height of each rectangle is given as $f\left(y_k^{*}\right)$, where${y_k}^{*}$ is the midpoint of each interval$\left[y_{k-1},y_k\right]$.

Use the above-given intervals to find the different values of $y_k^{*}$.

$\begin{array}{rcl}{y}_1^{*}& =& \frac{0+1}{2}=0.5\\ y_2^{*}& =& \frac{1+2}{2}=1.5\\ y_3^{*}& =& \frac{2+3}{2}=2.5\\ y_4^{*}& =& \frac{3+4}{2}=3.5\end{array}$

Part (c) Step 1: Given information.

In between ranges $y=0$ and $y=4$, the function is $f\left(x\right)=x^2$.

The goal is to use a midpoint sum with four rectangles to approximate the area of this region using a Riemann sum.

Part (c) Step 2: Calculation

Use the definition of function$f\left(x\right)=x^2$ to determine the inverse function, $f^{-1}\left(x\right)$.

$\begin{array}{rcl}f\left(x\right)& =& x^2\\ y& =& x^2\\ \sqrt{y}& =& x\\ f^{-1}\left(x\right)& =& \sqrt{x}\end{array}$

To calculate the values of the inverse function at these places, use the midpoint values.

$\begin{array}{rcl}{f}^{-1}(0.5)& =& \sqrt{0.5}\\ & =& 0.71\end{array}$

Use the next midpoint value to determine the next value of function.

$\begin{array}{rcl}{f}^{-1}(1.5)& =& \sqrt{1.5}\\ & =& 1.22\end{array}$

Use the next midpoint value to determine the next value of function.

$\begin{array}{rcl}{f}^{-1}(2.5)& =& \sqrt{2.5}\\ & =& 1.58\end{array}$

Use the next midpoint value to determine the next value of the function.

$\begin{array}{rcl}{f}^{-1}(3.5)& =& \sqrt{3.5}\\ & =& 1.87\end{array}$

Part (d) Step 1: Calculation of the required result.

 Each disk's area is specified as$A_k=\pi ·{\left(f^{-1}\left(y_k^{*}\right)\right)}^2$.

A disk's volume is indicated as$V_k=A_k·\Delta y$.

The two equations are combined to provide the following final volume equation:

$\begin{array}{rcl}{V}_k& =& A_k·\Delta y\\ & =& \pi ·{\left(f^{-1}\left(y_k^{*}\right)\right)}^2·\Delta y\end{array}$

Calculate the volume of the second disc using the value $k=2$ and the aforementioned parameters.

$\begin{array}{rcl}{V}_2& =& A_2·\Delta y\\ & =& \pi ·{\left(f^{-1}\left({y_2}^{*}\right)\right)}^2·\left(1\right)\\ & =& \pi ·{\left(f^{-1}(1.5)\right)}^2\\ & =& \pi ·(1.22{)}^2\\ & =& 1.5\pi \end{array}$