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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-4

Quadric Surfaces — Question 5

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Question 5

The surface z=x2+4y2−2x+8y+7z=x^2+4y^2-2x+8y+7 is cut by the plane z=10z=10. Identify the surface, locate its vertex, and find the semiaxes and area of the resulting planar region.

Original worksheet page 1: question and worked solution for 6-4-005
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Question 5 – Solution

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

See the diagram in the original worksheet below.

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

Cut at z=10z=10 The boundary satisfies (x−1)2+4(y+1)2=8(x-1)^2+4(y+1)^2=8, or (x−1)28+(y+1)22=1.\frac{(x-1)^2}{8}+\frac{(y+1)^2}{2}=1. Its semiaxes are 222\sqrt 2 and 2\sqrt 2, so the enclosed area is π(22)(2)=4π\boxed{\pi(2\sqrt 2)(\sqrt 2)=4\pi}.

Original worksheet page 2: question and worked solution for 6-4-005

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