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Volume of polyhedron consisting of non-linear curves .
Related problem ( S1 ) :
Let T/ PQRS denotes a pyramid where PQRS is a square with
each side 1/2 unit . TP = 1/4 un. being perpendicular to the base .
Both TQ and TS are straight lines while TR is a quadratic curve
with equation z = 1/4 ( 1-2x) ( 1-2y) where PQ denotes the x-axis ,
PS denotes the y-axis and PT denotes the z-axis .
How to find the volume of the pyramid ?
( Perhaps we can cut the pyramid into 2 halves at TPR to get
2 pyramids S/ TPR and Q/TPR , but how to find the area of the
base TPR ? )
Related problem (III)
Inside a triangle E 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 similar triangle A , 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 E , find the probability that the
point lies inside A and X at the same time .
Hi bob bundy ,
Many thanks again !
Now let us cut the polyhedron into 2 layers along VMXNZLV ,
the lower part is a triangular prism with height 1/6 un and area 1/2* 1/2* 1/2
= 1/8 sq un . Thus its vol = 1/6 * 1/8 = 1/48 cu un .
We shall denote the upper part by H , which can further be dissected into 4 portions .
The 1st one being a pyramid , V/WKLM , will be denoted by H1 .
The 2nd one also a pyramid , Z/KYNL , will be denoted by H2 .
The 3rd one also a pyramid , X/ WYNM , will be denoted by H3 .
The 4th portion remained is a prism , LMN/KWY, will be denoted by H4 .
The height of H1 is found to be √1/32 = 1/2 * √1/8 un , while the base
is a rectangle with area 1/8 * √1/8 sq un , thus vol of H1 = 1/3 * 1/2 * √1/8 *1/8 * √1/8
= 1/6 * 1/8 * 1/8 = 1/384 cu un
The height of H2 is 1/4 un , while the base is also a rectangle with area 1/4 * 1/8 sq un ,
thus vol of H2 = 1/3 * 1/4 * 1/4 * 1/8 = 1/384 cu un
The height and area of H3 is the same as H2 , thus the vol of H3 = 1/384 cu un .
The height of H4 is 1/8 un , while the base being a triangle with area 1/2 * 1/4 * 1/4
= 1/ 32 sq un , thus the vol of H4 = 1/ 256 cu un .
Thus the total volume of the polyhedron = 1/48 + 1/384 + 1/384 + 1/384 + 1/256
= 1/48 + 1/128 + 1/256 = 16/768 + 6/768 + 3/768 = 25/ 768 cu un .
Divided by the area of the base of the polyhedron , we also get its average height
= (25/768)/ (1/8) = 25 / 96 un . ( about 0.26 ) while the square of 0.26 is about 0.0677 .
Hi tacowreck ,
What result did you get for the volume ?
Hi bob bundy ,
Here is another problem . Before I have learnt the Geogebra , please help me once more !
Problem ( 3 T ) :
Let PQRSTUP / VWXYZKV denotes a polyhedron with
rt. ∟. Δ PRT as base where PR = PT = 1/2 unit Q, S and U are mid-points of PR , RT and TP respectively .VP , WQ , XR ,YS , ZT and KU are all ⊥ to Δ PRT , with VP = XR = ZT = 1/6 un. and WQ = YS = KU
= 7/24 un. WYK forms a triangle.
Find the volume of the polyhedron .
Hi bobbym ,
That's right !
Let us return to #1 . If the condition " keep parallel " is
not necessary , will the result be P = 1/4 * 1/4 = 1/16 ?
Hi bobbym ,
You are right ! For the case of perpendicular , the
answer is intuitively to be 1/4 .
For them to be parallel , the answer will be just the same
as the problem involving 2 segments .(i.e. P = 1/3 )
Hi bobbym ,
For the 2 smaller rectangles ( not triangles ) to be parallel ,
I mean that they are in the position like a " = " sign .
For they to be perpendicular , I mean that they are in the
position like a " τ " sign .
Thus in your diagram they should be mutually perpendicular .
Hi bob bundy ,
You are right , XVPZX is a rhombus , in addition ,
both XVZ and VPZ are equilateral Δs .
Let (i) H1 denotes the pyramid P/QUZV ;
(ii) H2 denotes the pyramid X/VZYW and
(iii) H3 denotes the prism QUTR/VZYW
For (i) , height = 1/2 *√1/8 (= √1/32 un.) ,
area of base = √1/8 * 1/4 sq . un .
Thus vol. of H1 = 1/3 * 1/2* √1/8 * √1/8 * 1/4
= 1/192 cu.un.
For (ii) , height = 1/4 un. , area of base = (√1/8 +√1/2)/2 * 1/2 *√1/8
= 3/32 sq.un.
Thus vol. of H2 = 1/3 * 1/4* 3 * √1/8 * 1/2*1/2 *√1/8
= 1/128 cu un .
For (iii) , height = 1/4 un , area of base = 3/32 sq un .
Thus vol. of H3 = 1/4 * 3/32 = 3/128 cu un
To sum up , volume of H = 1/192 + 1/128 + 3/128
= 7/192 cu.un (about 0.036) [ exactly the same as Problem (II) ]
Hi bob bundy ,
Another problem is waiting ! Thanks in advance .
Problem (III)
Let H denotes a polyhedron PQRSTUP/VWXYZ where the base is a
rt. ∟. Δ PRT with PR = PT = 1/2 unit . Q, S and U are
mid-points of PR , RT and TP respectively .VQ, WR ,XS ,YT and
ZU are all ⊥ to Δ PRT , with VQ = WR = YT = ZU = 1/4 un.
and XS = 1/2 un. ; X and Z , Z and P , P and V , V and X are
joined . Notice that XP and ZV are co-planed . XZPV seems to be a
kite with XZ = XV and ZP = PV . ( While XZTY in Problem (II) should be
a parallelogram with unequal adjacent sides .)
Find the volume of H .
(i) For H1 , area of the base = 1/4* 1/4 = 1/16 sq. unit while
height = 1/4 un , thus volume of H1 = 1/3* 1/4 * 1/16 = 1/192 cu.un
(ii) For H2 , area of the base = ( 1/4 + 1/2 )/2 * 1/4 = 3/32 sq.un while
height = 1/4 un , thus volume of H2 = 1/3 * 1/4 * 3/32 = 1/128 cu.un
(iii) For H3 , area of the base = 3/32 sq.un while height = 1/4 un ,
thus volume of H3 = 1/4 * 3/32 = 3/128 cu. un
To sum up , volume of H = 1/192 + 1/128 + 3/128 = 7/192 cu.un
(about 0.036)
Hi bob bundy ,
Your diagram is indeed what I want . Originally I expect you will
arrange the vertices in an anti-clockwise direction , and with the
pentagon PVXWR facing me . Now with your diagram I can look from
the back and obtain more integral impression of the polyhedron .
The whole polyhedron , as you said , can be divided into 3 portions :
(i) pyramid USYZ/T , will be denoted by H1 .
(ii) pyramid VZYW/X , will be denoted by H2 .
(iii) prism PUSR/VZYW , with cross - section being a trapezium , will be
denoted by H3 .
I shall calculate their volumes one by one .
Thanks again and I promise next time I will make the statement more
clearly .
Thanks bob bundy !
Here is another problem , wish you will help .
Problem (II)
Let H denotes a polyhedron PQRSTUVWXYZ where the base is a
rt. ∟. Δ PRT with PR = PT = 1/2 unit . Q, S and U are
mid-points of PR , RT and TP respectively .VP , XQ , WR ,YS and
ZU are all ⊥ to Δ PRT , with VP = WR = YS = ZU = 1/4 un. and
XQ = 1/2 un.
Find the volume of H .
Hi bob bundy ,
Thanks much to your analysis to the problem and the nice diagrams !
Let us consider the prism firstly . Since the area of Δ TPS = 1/2 *
1/2 * 1/8 sq.unit while the height = 1/2 unit , thus the volume
of the prism = 1/64 ( about 0.0156 ) cu.unit
Then for the pyramid , ST can be found to be √(17/64 ) unit ,
Thus the area of the base STVR = 1/2 * √(17/64 ) = (√17)/16 sq.unit .
But for the height , it is too bad for me to find its value with trigonometry for I have forgotten most of the trigonometry ! I have to find the volume of the pyramid by other means .
Let us cut the pyramid into 2 halves through Δ UYZ , we choose the
portion UYSTZ , also a pyramid , and denoted by H .
For H we shall cut it into 2 parts again through Δ USZ ,
part 1 , the pyramid YSUZ will be denoted by H1 , while
part 2 , the pyramid TSUZ , will be denoted by H2 .
For H1 , the point S will be considered as the new vertex with Δ UYZ as base . As SY being perpendicular to Δ UYZ ,
thus the height of H1 will be 1/4 unit .
Since the area of Δ YWU = 1/2 * 1/2 * 1/4 = 1/16 sq.unit
while the area of Δ YWZ = 1/32 sq.unit , thus the area of
Δ UYZ = 1/32 sq.unit also . ( In fact since YW ⊥ UW , the
area of Δ UYZ may be found directly to be 1/2 * 1/2 * 1/8 =
1/32 sq.unit .) Thus the volume of H1 = 1/3 * 1/4 * 1/32 =
1/384 cu. unit .
For H2 , the point U will be considered as the new vertex
with Δ STZ as base .
The height of H2 , UZ = 1/8 unit , while the area of Δ STZ =
1/2 * 1/4 * √(17/64 ) = (√17) / 64 sq. unit . Thus the volume of H2 = 1/3 * 1/8 * (√17) / 64 = (√17) / 1536 cu.unit .
So the volume of H = H1 + H2 = 1/384 + (√17) / 1536
= ( 4 + √17)/ 1536 cu.unit
Therefore the volume of the original pyramid = ( 4 + √17)/ 768 cu .unit
( about 0.01 cu.unit )
( Since the area of the original pyramid = (√17)/16 sq.unit . Thus the
height with U as vertex = 1/( 4√17) + ( 1/16 ) unit ( about 0.12 unit ) )
Lastly the volume of the original polyhedron will be found to be
1/64 + 0.01 = 0.016 + 0.01 = 0.026 cu.unit (approximately )
Hi bob bundy , thanks for your reply !
You should not join V to P , instead you should join V to R , and
also S to T . If you like you may also join U to S and U to R .
Really the point Y is not necessary in the problem . I am sorry
for my statement has not been clear enough , is it better to denote
the polyhedron by PQRSTUVQP ?
(I) Let X denotes a polyhedron with vertices PQRSTUV where
the base PQRS is a square with sides 1/2 unit . TP and VQ are
straight lines each with length 1/8 unit and perpendicular to PQRS .
W is the mid-point of PQ while Y is the mid-point of RS . UW with
length 1/4 unit is perpendicular to PQRS also . Z is the mid-point of
TV .( also of UW )
Find the volume of X .
Hi bobbym ,
The value 13/120 is quite near 1/9 = 13/ 117 .
Hi bobbym ,
Once you put either A or B into E , you cannot move it to any
position non-parallel to E , otherwise you just can't put it into E .
Thus A or B must be parallel to E .
In fact for a rectangle with 1 unit * x unit which can be put into E
in a non-parallel position ( with 1 side parallel to a diagonal of E ),
the greatest value of x is about 0.447 .
Related problem (II)
Inside a square E of 1*1 there are 2 rectangles A and B
both of 1* 1/2 and move freely inside E . If a point is chosen
randomly on E , find the probability that the point lies inside
A and B at the same time if
(i) A and B are mutually "parallel ".
(ii) A and B are mutually " perpendicular ".
Hi bobbym ,
I apologize that I had made mistakes again in the last post .
Under the condition of a2 = b2 , the probability required should
be 5/12 * 5/12 = 25/144 ( about 0.17)
I am not sure which answer provided by me is correct , or may
be all wrong .
It seems it is impossible to solve this problem by solid geometry ,
for there are too many variables thus out of the capacity of solid
geometry . Perhaps this problem may be solved by multiple integration
which I am not familiar . Thus I shall postpone to discuss this problem
for myself , but wish someone skilful in multiple integration may find a solution for it .
Later I shall discuss various related problems other than triangles .
Let us try to solve the original problem step by step from special case to general case .
Let (a1,a2) denotes the coordinate of VA and (b1,b2) denotes the coordinate of VB .
Suppose a2 = b2 , i.e A and B have same distance with the base of the large Δ .
The length of the base of the similar Δ formed by overlapping A and B
will be 1/2 - [ max ( a1,b1 ) - min ( a1,b1 )] i.e. 1/2 - |a1-b1| ( absolute value )
Since both a1 and b1 lie between 0 and 1/2 , thus |a1-b1| also lie between
0 and 1/2 . We can find the expected value of 1/2 - |a1-b1| by solid
geometry .
Let PQRST denotes a pyramid with base PQR being a Δ where PQ
= 1/2 unit ( length of |a1-b1| ) while PR = 1/2 unit ( range of a2 or b2 : also from
0 to 1/2 ) . RPTS being a square with length 1/2 unit , where PT denotes the value of
1/2 - |a1-b1| = 1/2 since |a1-b1| = 0 when both a1 = b1 = 0 .
Similarly RS = 1/2 when a1 = b1 = 0 with a2 and b2 = 1/2 .
To calculate the volume of the pyramid we treat the square as base with
Q as the vertex .
Since V = 1/3 * 1/2 * 1/2 * 1/2 = 1/24 cu.unit
Since the area of the original base Δ PQR = 1/2 * 1/2 * 1/2 = 1/8 sq.unit
Thus the average height ( expected value of 1/2 - |a1-b1| ) = 1/3 unit .
Thus the area of the similar Δ formed by overlapping A and B will be
1/3 * 1/3 * 1/2 sq.unit , dividing by the area of the large Δ being 1*1*1/2
sq.unit , we get the probability required = 1/9 .
Nextly we shall find what will be the result if a2 ≠ b2 .
Perhaps we can find the probability in another way .
In the diagram of #13 let the coordinates of the large Δ
denoted by (0,0) ( 1,0) and ( 0,1) , and let the 2 smaller Δs
denoted by A and B . Their vertices at right angle will be
represented by VA and VB . It is clear that both VA
and VB lie within the Δ (denoted by D) with coordinates (0,0) (1/2,0) and (0,1/2) or intersections of the 3 lines with equations
(i) y=0 ; (ii) x=0 and (iii) x+y = 1/2 .
If a point is chosen randomly inside D , what will be the expected
values of its coordinates ?
Hi bobbym ,
I made some mistakes in finding the volume of
the polyhedron in the past 3 posts . Now I shall
start again from the beginning and use your diagram
in #13 as reference .
Let X denotes a polyhedron with 6 vertices PQRSTU and 7 faces .
(1) base PQRS being a square with sides 1/2 unit .
(2) Δ PTS with PT = 1/3 unit where TP is perpendicular to the base .
(3) Δ PTQ being congruent to Δ PTS .
(4) Δ RUQ with RU = 1/2 unit where RU is perpendicular to the base .
(5) Δ RUS being congruent to Δ RUQ .
(6) Δ TQU .
(7) Δ TSU being congruent to Δ TQU .
In fact PT denotes the expectation of the length
of the common portion of the bases of the 2
smaller Δs ( the red and blue ) when the distance
of their bases with the base of the large Δ are
fixed to be 0 during moving in parallel direction ,
while RU denotes the expectation of the length
of the common portion of the bases of the 2
smaller Δs when the distance of their bases
with the base of the large Δ are fixed to be 1/2
unit .
By formula we find that PT = 1/3 unit while RU is
fixed to be 1/2 unit .
In this case UT will be a straight line ( linear )
otherwise it may be a curve of 2nd order if it
represents certain areas etc .
The polyhedron represents the distribution of data
of various expectation of the common portion of
the bases of the 2 smaller Δs according to the
various distances of their bases with the large Δ .
If we divide the volume of X by its base , we
get the overall expectation .
To find the volume of X , it may be easier to cut
X into 2 halfs at TU and get a pyramid Y .
The base of Y is a trapezium PRUT with PT // RU ,
since the height of the trapezium PR = √1/2 unit ,
∴ the area of PRUT may be found to be
( 1/3 + 1/2 ) / 2 * √1/2 = √1/2 * 5/12 sq.unit .
( If UT is a curve instead of a straight line , then the
area of PRUT should be found by integration if
we know its equation .)
Since trapezium PRUT and Δ PRS are mutually
perpendicular , thus the height of the pyramid is
in fact the height of the Δ SPR .
Let SW denotes the height meeting PR at W ,
in fact SW = PW = WR = 1/2 PR = 1/2 √1/2 unit .
Now we can calculate the volume of Y :
V = 1/3 * 1/2 √1/2 * √1/2 * 5/12 cu. unit
= 5/ 144 cu.unit
∴ V of X = 5/ 72 cu. unit
Thus the average height on PQRS
= ( 5/72)/ ( 1/4 ) unit
= 5/18 unit
The above value denotes the expectation of
the base of common portion ( also a similar
and parallel Δ ) of the 2 smaller Δs in # 13 .
Thus the probability required
= (5/18)*(5/18) = 25/324 ( about 0.077 )
Hi bobbym ,
In fact the problem of the polyhedron is much related
to that of the 3 Δs , perhaps we can obtain the
same result through different ways .
In polyhedron Y ,since trapezium PRUT and Δ PRS are mutually
perpendicular , thus the height of the pyramid is in fact the
height of Δ SPR .
Let SW denotes the height meeting PR at W ,
in fact SW = PW = WR = 1/2 PR = 1/2 √1/2 unit .
Now we can calculate the volume of Y :
V = 1/3 * 1/2 √1/2 * √1/2 * 5/48 cu. unit
= 5/ 576 cu.unit
Thus V of X = 5/ 288 cu. unit
Thus the average height ( expected area ) on PQRS
= ( 5/288)/ ( 1/4 ) sq.unit
= 5/72 sq.unit
The above value denotes the expectation of the common
portion of the 2 smaller Δs in # 13 , thus the probability
required = (5/72)/(1/2) = 5/36 ( about 0.14 )
Hi bobbym , wish you will have a smooth move !
Now let us continue to the polyhedron X :
If we cut X into 2 halfs at TU , we get 2 pyramids .
Let us choose the left one PRUTS , denoted by Y . We may
treat Y as a pyramid having 5 faces with PRUT ( being a trapezium ) as base , and S as vertice .
We may find that PR = √1/2 unit ; ST = √ 37/144 unit ;
SU = √ 17/64 unit and UT = √ 289/576 unit .
The area of base PRUT may also be found as
√1/2 * 5/48 sq.unit .
How to find the height of the pyramid Y so that we can
calculate the volume of Y ?