This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation
will generate a heart. I have tried and the following pictures show that fact.
This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation
will generate a heart. I have tried and the following pictures show that fact.
I just read the following result due to LossSloane published in J. Funct. Anal. this year [here]. This is just a lemma in their paper that I found very interesting.
Let be any function in . Consider the inversion and set
.
Then and
.
Proof. For fixed consider the regions
and likewise,
.
Let be points in . If we denote by the reflection point of with respect to the unit ball, i.e.
we then have the following wellknown identity
.
The proof of the above identity comes from the fact that
.
Indeed, by squaring both sides of
we arrive at
which is obviously true. Similarly, the last identity also holds. If we replace by we also have
.
Generally, if we consider the reflection point of over a ball , i.e.
we still have the fact
.
Indeed, one gets
.
Similarly,
.
Such identity is very useful. For example, in () the following holds
.
This type of formula has been considered before when here. For a general case, Lieb and Loss introduced another method in their book published by AMS in 2001. Here we introduce a completely new proof. At first, if by the potential theory, one easily gets
.
If , one needs to make use of the reflection point of and the above identity to go back to the first case. The point here is . The integral is obviously continuous as a function of . The above argument is due to professor X.X.W.
Let us consider the following integral in
when varies.
Obviously, the sphere can be parametrized as the following
so
If we wish to work on the average, the formula is much simpler than that, precisely
that means
where the bar means the average.
More general, we get
.
Today, we try to evaluate the following surface integral in . I found this result in a paper published in Math. Z. 198, 277289 (1988). This topic can be considered as a continued part to the following topic.
Proposition. Let be a continuous function. Then we have
here we denote .
Proof. Remark that the integral only depends on , so that we may assume and introduce the following spherical coordinate
where
.
This system of coordinates can be seen via the picture below
The solid common to two (or three) right circular cylinders of equal radii intersecting at right angles is called the Steinmetz solid. Two cylinders intersecting at right angles are called a bicylinder, and three intersecting cylinders a tricylinder. Half of a bicylinder is called a vault.

For two cylinders of radius oriented long the – and axes gives the equations
(1)

(2)

which can be solved for and gives the parametric equations of the edges of the solid,
(3)


(4)

The surface area can be found as , where
(5)


(6)

Taking the range of integration as a quarter or one face and then multiplying by 16 gives
(7)

The volume common to two cylinders was known to Archimedes (Heath 1953, Gardner 1962) and the Chinese mathematician Tsu Ch’ungChih (Kiang 1972), and does not require calculus to derive. Using calculus provides a simple derivation, however. Noting that the solid has a square cross section of sidehalflength , the volume is given by
(8)

(Moore 1974). The volume can also be found using cylindrical algebraic decomposition, which reduces the inequalities
(9)

to
(10)

giving the integral
(11)

If the two right cylinders are of different radii and with , then the volume common to them is
(12)

where is the complete elliptic integral of the first kind, is the complete elliptic integral of the second kind, and is the elliptic modulus.
The curves of intersection of two cylinders of radii and , shown above, are given by the parametric equations
(13)


(14)


(15)

(Gray 1997, p. 204).
The volume common to two elliptic cylinders
(16)

with is
(17)

where (Bowman 1961, p. 34).
For three cylinders of radii intersecting at right angles, The resulting solid has 12 curved faces. If tangent planes are drawn where the faces meet, the result is a rhombic dodecahedron (Wells 1991). The volume of intersection can be computed in a number of different ways,
(18)


(19)


(20)

(Moore 1974). According to the protagonist Christopher in the novel The Curious Incident of the Dog in the NightTime, “…People go on holidays to see new things and relax, but it wouldn’t make me relaxed and you can see new things by looking at earth under a microscope or drawing the shape of the solid made when 3 circular rods of equal thickness intersect at right angles” (Haddon 2003, p. 178), which is of course precisely the Steinmetz solid formed by three symmetrically placed cylinders.
Four cylinders can also be placed with axes along the lines joining the vertices of a tetrahedron with the centers on the opposite sides. The resulting solid of intersection has volume
(21)

and 24 curved faces analogous to a cubeoctahedron compound (Moore 1974, Wells 1991).
Six cylinders can be placed with axes parallel to the face diagonals of a cube. The resulting solid of intersection has volume
(22)

and 36 curved faces, 24 of which are kiteshaped and 12 of which are rhombic (Moore 1974).
Why is for being the unit sphere.
,
by rotational symmetry of the sphere. The integral is
.
And the easiest way to find (should you want to) is again by symmetry
,
hence .