Chapter 1.5, Problem 39E

(a)

To determine

To rewrite:the for the circumference of the circle.

Answer to Problem 39E

Thecircumference of the circleis rewritten as $r=\frac{c}{2\pi }$

Explanation of Solution

Given:

Circumference of the circle $c=2\pi r$

Calculation:

TheCircumference of the circle is rewritten as

  $\begin{array}{l}c=2\pi r\\ r=\frac{c}{2\pi }\end{array}$

Conclusion:

Therefore, thecircumference of the circle is rewritten as $r=\frac{c}{2\pi }$

(b)

To determine

To compute: the radius for the given circumference.

Answer to Problem 39E

Theradius is $r_1=1.1,r={1.3}_2,r_3=1.4$

Explanation of Solution

Calculation:

Circumference of the circle $c=2\pi r$

  $\begin{array}{l}{r}_1=\frac{c_1}{2\pi }=\frac{7}{\mathrm{2.3.14}}=1.1\\ r_2=\frac{c_2}{2\pi }=\frac{8}{\mathrm{2.3.14}}=1.3\\ r_3=\frac{c_3}{2\pi }=\frac{9}{\mathrm{2.3.14}}=1.4\\ r_1=1.1,r={1.3}_2,r_3=1.4\end{array}$

Conclusion:

Therefore, theradius is $r_1=1.1,r={1.3}_2,r_3=1.4$

(c)

To determine

To find: the area of cross section in terms of circumference.

Answer to Problem 39E

Thearea of cross section in terms of circumferenceis $S=\frac{c^2}{4\pi }$ .

Explanation of Solution

Calculation:

Thearea of cross section in terms of circumference iscalculated below

  $\begin{array}{l}S=\pi r^2\\ $ S=\pi .{(\frac{c}{2\pi })}^2$S=\frac{c^2}{4\pi }\end{array}$

Conclusion:

Therefore, thearea of cross section in terms of circumference is $S=\frac{c^2}{4\pi }$ .