Chapter 14: Q 5 (page 1149)

Give an intuitive explanation of why the paraboloid $z = x^{2} + y^{2}$ is smooth but the surface of the rectangular solid bounded by the coordinate planes and the planes $x = 14, y = 5, and z = 32$ is only piecewise smooth.

Short Answer

Expert verified

The figure is as below:

Step by step solution

Given Information

It is given that $z = x^{2} + y^{2}$ ,

$x = 14, y = 5, z = 32$ is piecewise smooth.

Explanation

There is always well defined choice of vector on surface of paraboloid. Hence given paraboloid is smooth.
Solid which is a rectangle is piecewise smooth, intersection of these forms corners which are sharp. Hence, these corners have not well defined normal vectors.
So, the rectangular solid that is bounded by coordinate planes and given planes $x = 14, y = 5, z = 32$ is piecewise smooth.

Figure

The figure is as below:

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Give an intuitive explanation of why the paraboloid $z = x^{2} + y^{2}$ is smooth but the surface of the rectangular solid bounded by the coordinate planes and the planes $x = 14, y = 5, and z = 32$ is only piecewise smooth.

Short Answer

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Give an intuitive explanation of why the paraboloidz=x2+y2 is smooth but the surface of the rectangular solid bounded by the coordinate planes and the planes x=14,y=5,andz=32is only piecewise smooth.

Short Answer

Step by step solution

Given Information

Explanation

Figure

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Give an intuitive explanation of why the paraboloid $z = x^{2} + y^{2}$ is smooth but the surface of the rectangular solid bounded by the coordinate planes and the planes $x = 14, y = 5, and z = 32$ is only piecewise smooth.