Volume of irregular polyhedron

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mr.Wong

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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 )
 
mr.Wong said:
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 withcoordinate ( 1/6 , 1/6 , 1/12 ) . Find the volume of the polyhedron H . ( looks somewhat like a tent )
Click to expand...


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Hi Subhotosh Khan ,

Thanks for your reply ! I think that this problem may be
solved directly by using double integration which I am
not familiar , so I need help .
 
mr.Wong said:
Hi Subhotosh Khan ,

Thanks for your reply ! I think that this problem may be
solved directly by using double integration which I am
not familiar , so I need help .
Click to expand...

Since it is a "Volume" problem. you may need triple integration.

If you are not familiar with those - why were you expected to solve this problem!

First step to solve the problem would be to approximately sketch the volume in question. Then you will need to define the limits of each part of integration.

What are the orientations of PQ and QR relative to x-y axes?

You may want to do an internet search of "volume integration" - come back and tell us what principle can be applied to this problem.
 
Last edited by a moderator:
Hi Subhotosh Khan ,

Thanks again ! The coordinates of P, Q, and R are (0, 1/2, 0), (0, 0, 0), and (1/2, 0, 0). Perhaps we have to integrate the x-axis from 0 to 1/2; the y-axis also from 0 to 1/2 and the z-axis from 0 to 1/12.

I had search "volume integration" from the web but I don't think I can
handle those principles since my calculus is too weak. Previously I attempt to solve the problem without integration, but now it seems impossible .
 
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