Today's calculation of Integral 501

kunny

Find the volume of the uion $A\cup B\cup C$ of the three subsets $A,\ B,\ C$ in $xyz$ space such that:

$A=\{(x,\ y,\ z)\ |\ |x|\leq 1,\ y^2+z^2\leq 1\}$
$B=\{(x,\ y,\ z)\ |\ |y|\leq 1,\ z^2+x^2\leq 1\}$
$C=\{(x,\ y,\ z)\ |\ |z|\leq 1,\ x^2+y^2\leq 1\}$

**Posted:** Tue Nov 03, 2009 9:05 am

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Today's calculation of Integral Digest

kenn4000

if you add eight chunks of equal size you can make $A\cup B \cup C$ into a sphere of radius $\sqrt 2$

so whats missing?

$\int_1^{\sqrt 2} \pi(2-\sqrt x)^2dx= \pi \int_1^ {\sqrt 2} 4-4\sqrt x +x dx$

$=\pi (4(\sqrt 2-1)-\frac{8}{3} (2^{\frac{3}{4}}-1)+\frac{1}{2})$

eight of these totals to $\frac{\pi}{3}(96\sqrt{2}-64\cdot 2^{\frac{3}{4}}-20)$

so the area of the shape is.. $\frac{4\pi}{3}2^{\frac{3}{2}}-\frac{\pi}{3}(96\sqrt{2}-64\cdot 2^{\frac{3}{4}}-20)$

but thats like.. 3.7.. which is less than the unit sphere.. which is clearly a subset.. something went wrong

**Posted:** Wed Nov 04, 2009 12:16 am

atomicwedgie

This is not the correct answer, since clearly the result must be greater than $2\pi$ . See the next post.

It is easier to first find the intersection $I = A \cap B \cap C$ then compute $|A \cup B \cup C| = |A| + |B| + |C| - 3|I|$ .

the solid $I$ looks like a rhombic dodecahedron with curved faces. Six fourfold vertices are located at at the points $(\pm 1, 0, 0)$ , $(0, \pm 1, 0)$ , and $(0, 0, \pm 1)$ . Eight threefold vertices are located $(\pm 2^{ - 1/2}, \pm 2^{ - 1/2}, \pm 2^{ - 1/2})$ .

Thus $I$ is composed of a cube centered at the origin with faces parallel to the axes with edge length $\sqrt {2}$ , and six rounded "pyramidal caps" each of whose volumes is further subdivided in eight congruent parts.

We calculate the volume of one of these parts:

$|V| = \int_{x = 0}^{1/\sqrt {2}} \int_{y = 0}^x \sqrt {1 - x^2} - \frac {1}{\sqrt {2}} \, dy \, dx$

$= - \frac {1}{4\sqrt {2}} + \int_{x = 0}^{1/\sqrt {2}} \int_{y = 0}^x \sqrt {1 - x^2} \, dy \, dx$

$= - \frac {1}{4\sqrt {2}} + \int_{x = 0}^{1/\sqrt {2}} x \sqrt {1 - x^2} \, dx$

$= - \frac {1}{4\sqrt {2}} - \frac {1}{2} \int_{u = 1}^{1/2} \sqrt {u} \, du$

$= - \frac {1}{4\sqrt {2}} + \frac {1}{2} \left[ \frac {2}{3} u^{3/2} \right]_{u = 1/2}^1$

$= - \frac {1}{4\sqrt {2}} + \frac {1}{3} \left(1 - \frac {1}{2\sqrt {2}}\right) = \frac {8 - 5\sqrt {2}}{24}$ .

Hence

$|I| = 48 |V| + 2 \sqrt {2} = 2(8 - 5\sqrt {2}) + 2\sqrt {2} = 16 - 8\sqrt {2} = 8(2 - \sqrt {2})$ .

Since $|A| = |B| = |C| = 2\pi$ , it follows that $|A \cup B \cup C| = 6(\pi - 4(2 - \sqrt {2}))$ .

atomicwedgie

I think I made an error in my assumption about how the union $A \cup B \cup C$ is constructed.

Instead of using the previous approach, we will calculate the volume $T$ enclosed in the $( + , + , + )$ octant enclosed by the surfaces $z = 1$ , $y^2 + z^2 = 1$ , $x^2 + y^2 = 1$ , and $x = y$ .

$|T| = \int_{y = 0}^{1/\sqrt {2}} \int_{x = y}^{\sqrt {1 - y^2}} 1 - \sqrt {1 - y^2} \, dx \, dy$

$= \int_{y = 0}^{1/\sqrt {2}} (1 + y)\sqrt {1 - y^2} + y^2 - y - 1 \, dy$

$= \int_{y = 0}^{1/\sqrt {2}} \sqrt {1 - y^2} \, dy - \frac {1}{2}\int_{u = 1}^{1/2} \sqrt {u} \, du + \left[\frac {y^3}{3} - ...$

$= \frac {\pi}{8} - \frac {1}{\sqrt {2}} + \frac {1}{3}$ .

32 of these, plus the volume of a single cylinder, gives the desired result:

$6\pi - 16\sqrt {2} + \frac {32}{3}$ .