Quadric Surfaces — Question 5

Question 5

The surface $z=x^2+4y^2-2x+8y+7$ is cut by the plane $z=10$. Identify the surface, locate its vertex, and find the semiaxes and area of the resulting planar region.

Question 5 – Solution

Strategy Complete squares in $x,y$; a horizontal cut of an elliptic paraboloid is an ellipse.

Standard form We obtain $$z=(x-1)^2+4(y+1)^2+2.$$ Thus the surface is an elliptic paraboloid opening upward with vertex $\boxed{(1,-1,2)}$.

Cut at $z=10$ The boundary satisfies $(x-1)^2+4(y+1)^2=8$, or $$\frac{(x-1)^2}{8}+\frac{(y+1)^2}{2}=1.$$ Its semiaxes are $2\sqrt 2$ and $\sqrt 2$, so the enclosed area is $\boxed{\pi(2\sqrt 2)(\sqrt 2)=4\pi}$.