# Cylindrical coordinates

## Homework Statement

Identify and sketch the graph of the surface y^2 + z^2 = 1. Show atleast one contour perpendicular to each coordinate axis

## The Attempt at a Solution

for the yz plane z = (1-y^2)^1/2 a circle of radius 2 centered at the origin

xy, set z=0 yields y = +/- (1)^1/2 which is a straight line at y+ and y-

xz, set y = 0 yields z = +/- 1^1/2 which is a straight line at z+ & z-

i do not get what the 3d model should look like

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Mark44
Mentor

## Homework Statement

Identify and sketch the graph of the surface y^2 + z^2 = 1. Show atleast one contour perpendicular to each coordinate axis

## The Attempt at a Solution

for the yz plane z = (1-y^2)^1/2 a circle of radius 2 centered at the origin
No. For the y-z plane (IOW, x = 0), you have y2 + z2 = 1, which is a circle of radius 1, centered at the origin.

Solving for z was really a wasted effort, since this is the equation of a circle.

xy, set z=0 yields y = +/- (1)^1/2 which is a straight line at y+ and y-
What does "straight line at y+ and y-" mean? In the x-y plane you have y = +/- 1, two horizontal lines
xz, set y = 0 yields z = +/- 1^1/2 which is a straight line at z+ & z-
Same comment as above.
i do not get what the 3d model should look like
What you need are more cross sections, such as the cross section in the planes x = 1, x = 5, x = 10, x = -1, x = -5, and x = -10. After you understand why I picked these cross sections, and graph a few of them, you'll have a good idea of what this surface looks like.

but there is no x value in the function.

Mark44
Mentor
but there is no x value in the function.
Which means that it is arbitrary; you can choose any value for x.

A clue that you're dealing in three dimensions is that they asked for the surface. If the problem had asked you to graph the curve y2 + z2= 1, you would have needed only two dimensions.