Volume of irregular polyhedron

Question

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 )

Answer 1

I take it the sides and bottom are flat but the top looks something like this:

I think you're just supposed to integrate the height over the triangular base, that is $$\begin{align}V&=\int_0^{\frac12}\int_0^{\frac12-x}\left[\frac12x-x^2+\frac12y-y^2-xy\right]dy\,dx\\ &=\int_0^{\frac12}\left[\frac12xy-x^2y+\frac14y^2-\frac13y^3-\frac12xy^2\right]_0^{\frac12-x}dx\\ &=\int_0^{\frac12}\left[\frac12x\left(\frac12-x\right)-x^2\left(\frac12-x\right)+\frac14\left(\frac12-x\right)^2-\frac13\left(\frac12-x\right)^3-\frac12x\left(\frac12-x\right)^2\right]dx\\ &=\left[-\frac14x\left(\frac12-x\right)^2-\frac1{12}\left(\frac12-x\right)^3+\frac12x^2\left(\frac12-x\right)^2+\frac13x\left(\frac12-x\right)^3\right.\\ &\left.+\frac1{12}\left(\frac12-x\right)^4-\frac1{12}\left(\frac12-x\right)^3+\frac1{12}\left(\frac12-x\right)^4+\frac16x\left(\frac12-x\right)^3+\frac1{24}\left(\frac12-x\right)^4\right]_0^{\frac12}dx\\ &=0+\frac1{96}+0+0-\frac1{192}+\frac1{96}-\frac1{192}+0-\frac1{384}=\frac1{128}\end{align}$$