SOLUTION TO PROBLEM H-11

Figure 1

Consider the figure above where the origin is at the center of the cube. A typical cross-section through our solid, perpendicular to the x-axis looks as in Figure 2.

Figure 2

Thus the area of the cross section will be equal to the area of the disk of radius 1/2 i.e. to $\pi /4$ plus the areas of the 4 triangle-like regions situated outside that disk. Let us calculate now the area of one of these regions. To that aim assume that for an arbitrary x between 0 and 1/2 we drew a plane through (x,0,0), perpendicular to the x-axis, producing the picture in Figure 2. Let's calculate the length of the horizontal side of one of these triangle-like regions. Using the Pythagorean theorem one gets that the aforementioned side equals $\sqrt{1-4x^2}/2$. Thus the area of each of the 4 triangle-like regions is

$$\begin{array}{l} \int_0^{\sqrt{1-4x^2}/2}\left( \frac{1}{2}-\... ...{1}{8}\arcsin\left(2\sqrt{\frac{1}{4}-x^2}\right). \end{array}$$

To find the volume we will integrate dx the area of the cross-section in Figure 2 for x between 0 and 1/2 and multiply it by 2. Thus the volume is

$$\begin{array}{l} V=\\ 2\int_0^{1/2}\left(\frac{\pi }{4}+4\le... ...right) \right)\, dx\\ =\frac{\pi}{2}-\frac{2}{3}. \end{array}$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Apr 5 20:24:55 CST 2002