# Solutions to Problem Set 9

Math 311-01/02

2-8-2018

[Parametrizing surfaces problems]

1. (a) Parametrize the surface by setting x and y to parameters u and v.

Then sketch the result.

(b) Parametrize the surface by setting and .

Then sketch the result.

(a)

(b) When and ,

The parametrization is

This problem shows that parametrizing a surface in different ways can lead to different-looking pictures. The parametrization in (b) is a polar coordinate parametrization; it's useful if you see combinations of squares of the variables.

2. Parametrize the surface by setting the independent variables x and z to parameters u and v.

Then sketch the result.

3. (a) Parametrize the curve by using a polar-coordinate parametrization with one parameter u.

(b) Using the result of (a), parametrize the cylinder . (Use the equations from (a), then set x to a second parameter v.)

Then sketch the result.

(a)

(b)

This problem shows that if a surface equation only involves 2 variables, you can parametrize it as a curve using one parameter, then set the remaining variable to the second parameter.

4. The point is one corner of a parallelogram. The sides emanating from P are given by the vectors and . Parametrize the parallelogram.

Then sketch the graph. (Use the ranges and for your parameters when you plot the graph.)

5. Parametrize the sphere

Then sketch the graph.

6. (a) Parametrize the curve .

(b) Parametrize the surface of revolution obtained by revolving the curve for , about the x-axis.

Then sketch the surface.

(c) Parametrize the surface of revolution obtained by revolving the curve for , about the y-axis.

Then sketch the surface.

(a)

(b)

(c)

7. Parametrize the surface which consists of the part of the cylinder satisfying .

Then sketch the surface.

I'll construct a segment from a point on the base of the cylinder straight up the side of the cylinder to the point on the plane.

can be parametrized by

A typical point on the base of the cylinder is .

If I go straight up the side of the cylinder to the plane , I get the point (since ).

The segment joining the two points is

Therefore, the surface may be parametrized by

The best way to escape from a problem is to solve it. - Brendan Francis

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