Volume of irregular polyhedron

[Guest]

Let H denotes a polyhedron with its base being a triangle
PQR with PQ = QR = 1/2 unit and angle Q = 90 degree .
The cover ( upper layer ) of the polyhedron is determined by
the equation : z= 1/2 x - x^2 + 1/2 y - y^2 - xy .
Thus for QR , the x-axis or y= 0 , the equation becomes
z = 1/2 x - x^2 being a quadratic curve over QR ;
for PQ , the y-axis or x = 0 , the equation becomes
z = 1/2 y - y^2 being a quadratic curve over PQ ;
for PR ( represented by x + y = 1/2 ) the equation becomes
z = 1/4 - x^2 - y^2 - xy also being a quadratic curve over PR .
We may find that the above 3 curves all have maximum
values = 1/16 unit at the mid-points of PQ ,QR and PR .
Moreover the polyhedron has an apex at the point with
coordinate ( 1/6 , 1/6 , 1/12 ) .
Find the volume of the polyhedron H . ( looks somewhat
like a tent )

[Guest]

1/128
[url]https://www.wolframalpha.com/input/?i=integrate%28integrate%28x%2F2-x^2%2By%2F2-y^2-xy,+{x,0,1%2F2-y}%29,{y,0,1%2F2}%29[/url]

Hope this helped,

R. Baber.

[Guest]

Thank you very much R.Baber !

You got an answer of 1/128 . I got an approximate answer of
a bit greater than 1/192 by myself , while from other place I got
an answer of 1/384 . So it seems your answer is most reliable .

[Guest]

R.Baber ,

Your answer of 1/128 is correct . Thanks again !

mr.wong

[Guest]

No problem it was just a simple case of integrating twice. For a fixed [tex]y[/tex] work out the area of a cross section (in the [tex]x[/tex] - [tex]z[/tex] plane) of your "polyhedron" ([tex]x[/tex] will vary from [tex]0[/tex] to [tex]1/2 - y[/tex])
[tex]\int_0^{\frac{1}{2} - y} \left(\frac{x}{2} - x^2 + \frac{y}{2} - y^2 - xy\right)dx[/tex]
[tex]= \left[\frac{x^2}{4} - \frac{x^3}{3} + \frac{x y}{2} - x y^2- \frac{x^2 y}{2}\right]_0^{\frac{1}{2}-y}[/tex]
[tex]= \frac{(\frac{1}{2}-y)^2}{4} - \frac{(\frac{1}{2}-y)^3}{3} + \frac{(\frac{1}{2}-y) y}{2} - \left(\frac{1}{2}-y\right) y^2- \frac{(\frac{1}{2}-y)^2 y}{2}[/tex]
[tex]= \frac{5 y^3}{6}-\frac{3 y^2}{4}+\frac{y}{8}+\frac{1}{48}[/tex]

Then add up (or more precisely integrate) all the areas of the cross sections to get the volume.
[tex]\int_0^{\frac{1}{2}} \left(\frac{5 y^3}{6}-\frac{3 y^2}{4}+\frac{y}{8}+\frac{1}{48}\right) dy[/tex]
[tex]= \left[\frac{5 y^4}{24}-\frac{y^3}{4}+\frac{y^2}{16}+\frac{y}{48}\right]_0^{\frac{1}{2}}[/tex]
[tex]= \frac{5 (\frac{1}{2})^4}{24}-\frac{(\frac{1}{2})^3}{4}+\frac{(\frac{1}{2})^2}{16}+\frac{(\frac{1}{2})}{48}[/tex]
[tex]= \frac{1}{128}[/tex]

Hope this helped,

R. Baber.

[Guest]

Thanks again R. Baber ,

In fact this problem is related to the following problem
concerning probability . Try to solve it if you are interested !

Inside a triangle E with vertices S , Q and T where SQ = QT = 1 unit
and angle Q = 90 degree , there is a smaller triangle X formed by joining
the mid-points of the 3 sides of E , and stays fixed in its position .
Another triangle A inside E , with lengths of sides being 1/2 of that
of E and parallel to E with corresponding vertices facing the same
direction , can move freely but parallelly inside E .
If a point is chosen randomly on X , find the probability that the
point also lies inside A .

mr.wong

[Guest]

If [tex]Q[/tex] is the origin, [tex]S[/tex] is [tex](1,0)[/tex], and [tex]T[/tex] is [tex](0,1)[/tex]. Then if the right angle of [tex]A[/tex] lies at [tex](x,y)[/tex] the overlap of [tex]A[/tex] and [tex]X[/tex] has an area of
[tex]z=\frac{x}{2} - x^2 + \frac{y}{2} - y^2 - xy[/tex]
and [tex](x,y)[/tex] lies in the triangle with vertices [tex](0,0)[/tex], [tex](0,1/2)[/tex], [tex](1/2,0)[/tex].

The probability of being in [tex]A[/tex] given [tex](x,y)[/tex] (assuming uniform distributions) is just (Overlap of [tex]A[/tex] and [tex]X[/tex])/(Area of [tex]X[/tex]) [tex]= 8z[/tex]. Just integrate over all possibilities of [tex](x,y)[/tex] to get the overall probability which is just [tex]8/128 = 1/16[/tex].

Hope this helped,

R. Baber.

[Guest]

Oops, sorry I made a mistake.
[tex]1/16[/tex] = the Probability times the area of choices of [tex](x,y)[/tex].
You need to divide by [tex]1/8[/tex], to get the right answer of [tex]1/2[/tex].

R. Baber.

[Guest]

Thanks R.Baber ,

Well done ! I got the answer of 1/2 by myself too.
But I would like to ask how do you know that the
overlap of A and X has an area of
z = 1/2 x - x^2 + 1/2 y - y^2 - xy ?

mr.wong

[Guest]

Just draw a diagram. You'll see that [tex]A[/tex] will cut [tex]X[/tex] in three places, the parts that get cut off are all right angled isosceles triangles. The upper left triangle has a side length of [tex]x[/tex], the lower right triangle has a side length of [tex]y[/tex], and the upper right triangle has a side length of [tex]1/2-x-y[/tex].
So the overlap has an area of [tex]\frac{(1/2)^2}{2} - \frac{x^2}{2} - \frac{y^2}{2} - \frac{(1/2-x-y)^2}{2}[/tex]
[tex]=\frac{x}{2}-x^2+\frac{y}{2}-y^2-xy[/tex]

Hope this helped,

R. Baber.

[Guest]

Thanks R.Baber !

There is a problem of the same type involving squares
which I shall post it in a new thread . I hope you will be
interested .

mr.wong