Chapter 8.3, Problem 109E

a.

To determine

To write: a formula in series notation that gives the surface area of the sphere flake.

Answer to Problem 109E

  $S={\displaystyle \sum _{n=1}^{\infty }4\pi }.$

Explanation of Solution

Given information: In a fractal , a geometric figure is repeated at smaller scales. The

Sphere flake shown is a computer-generated fractal that was created by Eric Haines. The radius of the large sphere is 1. Attached to the large sphere are nine sphere of radius 1/3. Attached to each of the smaller sphere are nine sphere of radius 1/9. This process is continued infinitely.

Calculation:

Surface area of the sphere is $4\pi r^2$ . The surface area of the sphere flakes is:

  $\begin{array}{c}S=4\pi {(}^1+9\left(4\pi {\left(\frac{1}{3}\right)}^2\right)+9^2\left(4\pi {\left(\frac{1}{9}\right)}^2\right)+\mathrm{....}\\ S=4\pi +4\pi +4\pi +\mathrm{........}={\displaystyle \sum _{n=1}^{\infty }4\pi }\end{array}$

b.

To determine

To write: a formula in series notation that gives the volume of the sphereflake.

Answer to Problem 109E

  $V={\displaystyle \sum _{n=1}^{\infty }\frac{4}{3}\pi {\left(\frac{1}{3}\right)}^n}.$

Explanation of Solution

Given information:

Calculation:

Volume of the sphere is $\frac{4}{3}\pi r^3$ . The surface area of the sphere flakes is:

  $\begin{array}{c}V=\frac{4}{3}\pi {(}^1+9\left(\frac{4}{3}\pi {\left(\frac{1}{3}\right)}^3\right)+9^2\left(\frac{4}{3}\pi {\left(\frac{1}{9}\right)}^3\right)+\mathrm{....}\\ V=\frac{4}{3}\pi +\frac{4}{3}\pi \left(\frac{1}{3}\right)+\frac{4}{3}\pi {\left(\frac{1}{3}\right)}^2+\mathrm{........}={\displaystyle \sum _{n=1}^{\infty }\frac{4}{3}\pi {\left(\frac{1}{3}\right)}^n}.\end{array}$

c.

To determine

To find: the surface area of the sphereflake finite or infinite, the volume is finite or infinite, if either is finite, find the value.

Answer to Problem 109E

The surface area is infinite and the volume is infinite.

  $V=2\pi \text{cubic}\text{units}.$

Explanation of Solution

Given information:

Calculation:

The surface area is infinite and the volume is infinite.

  $V=\frac{\frac{4}{3}\pi }{1-\frac{1}{3}}=2\pi \text{cubic}\text{units}.$