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Calculus IIIChapter 1: 3-Dimensional SpaceSection 1-4

Quadric Surfaces — Question 6

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Question 6

Find and describe the intersection of x2+y2+z2=25andz2=x2+y2.x^2+y^2+z^2=25\quad\text{and}\quad z^2=x^2+y^2.

Tasks

  1. Determine every component of the intersection.

  2. Find each component’s exact height and radius.

  3. Verify the result in both original equations.

Original worksheet page 1: question and worked solution for 1-4-006
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Question 6 – Solution

Strategy Substitute the cone relation into the sphere, then retain both signs of zz.

See the diagram in the original worksheet below.

Step 1: Solve for height Since x2+y2=z2x^2+y^2=z^2, the sphere gives z2+z2=25⇒2z2=25⇒z=±52.z^2+z^2=25\Longrightarrow 2z^2=25\Longrightarrow z=\pm\frac 5{\sqrt 2}.

Step 2: Solve for radius On either level, x2+y2=z2=252.x^2+y^2=z^2=\frac{25}{2}. Thus the intersection consists of two circles: z=±52,x2+y2=252.\boxed{z=\pm\frac 5{\sqrt 2},\qquad x^2+y^2=\frac{25}{2}}. Each has radius 5/2\boxed{5/\sqrt 2}. Both signs are essential because the cone is double. Substitution verifies both circles in both original equations.

Original worksheet page 2: question and worked solution for 1-4-006

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