Question

Let R = [ 0 , 1 ] × [ 0 , 1 ] R=[0,1]×[0,1]. Find the volume of the region above R R and below the plane which passes through the three points ( 0 , 0 , 1 ) (0,0,1), ( 1 , 0 , 8 ) (1,0,8) and ( 0 , 1 , 9

Answer 1

The three vectors $\langle0,0,1\rangle$, $\langle1,0,8\rangle$, and $\langle0,1,9\rangle$ each terminate on the plane. We can get two vectors that lie on the plane itself (or rather, point in the same direction as vectors that do lie on the plane) by taking the vector difference of any two of these. For instance,

$\langle1,0,8\rangle-\langle0,0,1\rangle=\langle1,0,7\rangle$

$\langle0,1,9\rangle-\langle0,0,1\rangle=\langle0,1,8\rangle$

Then the cross product of these two results is normal to the plane:

$\langle1,0,7\rangle\times\langle0,1,8\rangle=\langle-7,-8,1\rangle$

Let $(x,y,z)$ be a point on the plane. Then the vector connecting $(x,y,z)$ to a known point on the plane, say (0, 0, 1), is orthogonal to the normal vector above, so that

$\langle-7,-8,1\rangle\cdot(\langle x,y,z\rangle-\langle0,0,1\rangle)=0$

which reduces to the equation of the plane,

$-7x-8y+z-1=0\implies z=7x+8y+1$

Let $z=f(x,y)$. Then the volume of the region above $R$ and below the plane is

$\displaystyle\int_0^1\int_0^1(7x+8y+1)\,\mathrm dx\,\mathrm dy=\boxed{\frac{17}2}$