Chapter 7.3, Problem 78E

(a)

To determine

The area of the cross sections corresponding to the given circumferences.

Answer to Problem 78E

The area of the cross sections has been find out.

Explanation of Solution

Given:

The height of the vase measured is $6\text{ inches }$ and the data of circumference is also given.

Calculation:

Circumference can be find out as;

  $\begin{array}{l}C=2\prod r\\ r=\frac{C}{2\prod }\end{array}$

Now, the formula to find out area is as follows:

  $\begin{array}{l}A=\prod r^2\\ $ A=\prod {(\frac{C}{2\prod })}^2$A=\frac{C^2}{4\prod }\end{array}$

S.no.	Height (in inches)	Circumference (in inches)	Radius (in inches)	Area ( in inches2)
0	0	5.4	0.8594	2.3205
1	0.5	4.5	0.7162	1.6114
2	1	4.4	0.7003	1.5406
3	1.5	5.1	0.8117	2.0698
4	2	6.3	1.0027	3.1584
5	2.5	7.8	1.2414	4.8415
6	3	9.4	1.4961	7.0315
7	3.5	10.8	1.7189	9.2819
8	4	11.6	1.8462	10.7079
9	4.5	11.6	1.8462	10.7079
10	5	10.8	1.7189	9.2819
11	5.5	9	1.4324	6.4458
12	6	6.3	1.0027	3.1584

Conclusion:

The area of cross sections has been find out.

(b)

To determine

The volume of a vase.

Answer to Problem 78E

The volume of a vase is ${\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}$

Explanation of Solution

Given:

n=12

Calculation:

The volume of an element of a vase can be find out as:

  $dV=A(y)dy$

Now, to calculate the volume of a vase can be obtained by integrating the volume of an element.

  $\begin{array}{l}{\displaystyle \int dV}={\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}\\ V={\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}\end{array}$

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f(y)dy=\frac{\Delta y}{2}(f(y_0)+2f(y_1)+2f(y_2)+2f(y_3)+2f(y_4)+\mathrm{.......}+2f(y_n-1)+f(y_n))}\\ {\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}=\frac{0.5}{2}(2.3205+2(1.6114+1.5406+2.0698+3.1584+4.8415+7.0315+9.2819+10.7079+10.7079+9.2819+6.4458)+3.1584)\end{array}$

Hence, Volume can be find out as:

  $V={\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}=37.7091{\text{ in}}^3$

Conclusion:

The volume of a vase is ${\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}$ .

(c)

To determine

The value of integral using trapezoidal rule.

Answer to Problem 78E

The approximate volume using trapezoidal rule is $37.7091{\text{ in}}^3$ .

Explanation of Solution

Given:

The interval is given $\left[0,6\right]$

Calculation:

The volume of an element of a vase can be find out as:

  $dV=A(y)dy$

Now, to calculate the volume of a vase can be obtained by integrating the volume of an element.

  $\begin{array}{l}{\displaystyle \int dV}={\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}\\ V={\displaystyle \underset{0}{\overset{6}{\int }}A(y)dy}\end{array}$

Conclusion:

The approximate volume using trapezoidal rule is $37.7091{\text{ in}}^3$ .