Math Problem Statement

Find a parametrization of the boundary curve ∂S with positive orientation for the given paraboloid surfaces and planes as described in the image.

Solution

Let us work through each question step by step.

Question 1

We are tasked with finding a parameterization for the boundary curve of S\partial S of the surface of the paraboloid z=6x2y2z = 6 - x^2 - y^2 above the plane z=3z = -3 with positive orientation.

Step 1: Understand the boundary curve

The paraboloid z=6x2y2z = 6 - x^2 - y^2 intersects the plane z=3z = -3. Setting z=3z = -3, 6x2y2=3    x2+y2=9.6 - x^2 - y^2 = -3 \implies x^2 + y^2 = 9. Thus, the boundary curve lies on the circle x2+y2=9x^2 + y^2 = 9 in the xyxy-plane.

Step 2: Parameterize the boundary

A circle of radius 9=3\sqrt{9} = 3 can be parameterized as: x(t)=3cos(t),y(t)=3sin(t),where t[0,2π].x(t) = 3 \cos(t), \quad y(t) = 3 \sin(t), \quad \text{where } t \in [0, 2\pi].

Step 3: Determine zz on the boundary

On the boundary, z=3z = -3. Hence, the parameterization becomes: S=3cos(t),3sin(t),3,t[0,2π].\partial S = \langle 3 \cos(t), 3 \sin(t), -3 \rangle, \quad t \in [0, 2\pi].

Final Answer for Question 1:

The correct choice is (c): 3cos(t),3sin(t),3\langle 3 \cos(t), 3 \sin(t), -3 \rangle.


Question 2

We are tasked with finding a parameterization for the boundary curve of S\partial S of the surface of the paraboloid z=2x2+2y24z = 2x^2 + 2y^2 - 4 below the plane z=2z = 2 with a normal vector pointing downward.

Step 1: Understand the boundary curve

The paraboloid z=2x2+2y24z = 2x^2 + 2y^2 - 4 intersects the plane z=2z = 2. Setting z=2z = 2, 2x2+2y24=2    x2+y2=3.2x^2 + 2y^2 - 4 = 2 \implies x^2 + y^2 = 3. Thus, the boundary curve lies on the circle x2+y2=3x^2 + y^2 = 3 in the xyxy-plane.

Step 2: Parameterize the boundary

A circle of radius 3\sqrt{3} can be parameterized as: x(t)=3cos(t),y(t)=3sin(t),where t[0,2π].x(t) = \sqrt{3} \cos(t), \quad y(t) = \sqrt{3} \sin(t), \quad \text{where } t \in [0, 2\pi].

Step 3: Determine zz on the boundary

On the boundary, z=2z = 2. Hence, the parameterization becomes: S=3cos(t),3sin(t),2,t[0,2π].\partial S = \langle \sqrt{3} \cos(t), \sqrt{3} \sin(t), 2 \rangle, \quad t \in [0, 2\pi].

Final Answer for Question 2:

The correct choice is (d): 3cos(t),3sin(t),2\langle \sqrt{3} \cos(t), \sqrt{3} \sin(t), 2 \rangle.


Summary of Answers:

  1. (c): 3cos(t),3sin(t),3\langle 3 \cos(t), 3 \sin(t), -3 \rangle.
  2. (d): 3cos(t),3sin(t),2\langle \sqrt{3} \cos(t), \sqrt{3} \sin(t), 2 \rangle.

Would you like detailed explanations on any specific part of the derivations?


Related Questions:

  1. How do we derive parameterizations for surfaces other than paraboloids?
  2. What is the role of the normal vector orientation in these problems?
  3. How would the parameterizations change if the paraboloids were rotated?
  4. Can the same method be applied to intersections with spherical surfaces?
  5. How does the parameterization ensure the correct boundary orientation?

Tip: Always confirm the boundary curve by substituting the parameterization into the original equations.

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Math Problem Analysis

Mathematical Concepts

Parametrization of curves
Intersection of surfaces
Circular motion parameterization

Formulas

x(t) = r * cos(t), y(t) = r * sin(t), z = constant
For paraboloids: z = a(x^2) + b(y^2) + constant

Theorems

Intersection theorem for surfaces
Properties of circles in parametric form

Suitable Grade Level

Undergraduate Level (Calculus III or Multivariable Calculus)