Question

I'm trying to figure out the shape I'll get if I rotate a line created by the intersection of two planes around the x-axis. It looks like it'd be a cylinder, but I don't think that's right. I've attached an image for reference.

Answer 1

Looks like you will get a frustum of a cone.

Answer 2

I am assuming the x-axis is as shown.

Answer 3

That is correct. I got as far as a cross product of the two vectors, but I'm not sure how to rotate it.

Answer 4

Ah, and I just saw your cone picture after typing that. Ok, so in the form Ax+By+Cz=D, I need to create something that looks like a circle looking straight on from x, and the slope comes from the line in this image.

Answer 5

Maybe I just need dinner first. :)

Answer 6

Needs more clarification. In the first picture, is the plane on the right y = constant (that is parallel to the x-z plane)? Is the plane symmetric with respect to the x and z axis (that is, does it extend as much in the positive direction as the negative direction along x and z axes?) Does the line of intersection pass through the y-axis?

If the answer is yes to all of the above, then the solid of revolution will be two symmetrical frustums of cone back to back.

When rotated about the x-axis, the left tip of the line will trace out a circle with the equation: y^2 + z^2 = a^2 and the right tip will trace out a circle with the equation y^2 + z^2 = b^2.

If the line passes through the y-axis and the plane is symmetric, then "a" will equal "b" and it will be two symmetric frustums. While the "a" end is tracing out the upper part of the circle on the left, the "b" end will be tracing out the lower part of the circle at the right end as the line is rotated about the x-axis:

Answer 7

Sorry for the delayed response. There is a plane at y=(constant), and it intersects another plane at x=(another constant)*z. The line of intersection is what I'm rotating around the x-axis to get a formula of the form:\[Ax^{2}+By^{2}+Cz^{2}=1\]

Answer 8

i think it would be a cone

Answer 9

We need to find three points on \(Ax^{2}+By^{2}+Cz^{2}=1\) to determine the three constants. We can choose all three points on the line of intersection which will all have the y-value of 10. One point is (0, 10, 0) which when substituted above gives B = 1/100. You can choose two other points depending on the equation of the plane x = (what constant?) * z. y will be 10 for both those points. Substitute both points above and solve the simultaneous equation for A and C.

Answer 10

I'm not sure if I'm clear on how you got B=1/100? I have (0,10,0), which with \[y^{2}+z^{2}=A^{2}\]
gives \[10^{2}+0^{2}=A^{2}\]
\[A=10\]

Answer 11

I have y=10, x=6z.

Answer 12

So looking at the circle at (6,10,1) from x-axis, I get 36+1=r^2.
At (0,10,0), I get 100+1==r^2.
This makes me wonder...could I take the derivative to get the slope of r?

Answer 13

Oh, duh. I see what I was missing. :P Should be there in a minute.

Answer 14

\(Ax^{2}+By^{2}+Cz^{2}=1\) ------- (1)

The plane x = 6z intersects with the plane y = 10.
When z = 0, x = 0. Therefore, (0, 10, 0) is a point on the line of intersection.
When z = 1, x = 6. Therefore, (6, 10, 1) is a point on the line of intersection.

We should not choose a third point on the line of intersection because it would lead to a pair of dependent equations when we plug all three points into (1). Instead let us find the radius of the circle when the previous point (6, 10, 1) is rotated about the x-axis:
\(r^2 = y^2 + z^2 = 10^2 + 1^2 = 101. ~~r = \sqrt{101}\).
Therefore, when y = 0, \(z = \sqrt{101}\). Thus, the third point is \((6, 0, \sqrt{101})\).

Plug the three points into (1)
(0,10,0): 0 +100B + 0 = 1; B = 1/100
(6, 10, 1): 36A + 100B + C = 1; 36A + 1 + C = 1; 36A + C = 0
(6,0, sqrt(101)): 36A + 0 + 101C = 1; 36A + 101C = 1
36A + 101C = 1
36A + C = 0
subtract:
0 + 100C = 1
C = 1/100
A = -C / 36 = -1 / 3600

\(\Large -\frac{x^2}{3600} + \frac{y^2}{100} + \frac{z^2}{100} = 1\)
or
\(\large-x^2 + 36y^2 + 36z^2 = 3600\)

This surface is a one sheet hyperboloid.
The intersection of any plane y = constant or z = constant with this surface will be a hyperbola (much like two frustums of cone held back to back). The intersection of of any plane x = constant with this surface will be a circle.

Wolfram plot: http://www.wolframalpha.com/input/?i=plot+-x^2+%2B36y^2+%2B36z^2+%3D+3600

Answer 15

Oh wow, that gave me some grief. The problem I kept running into was precisely that I *was* using the same line (like you said not to). :)

The good news is, after a few hours and your generous help, I finally understand it! Thank you so much for your help. I was beginning to think I might never get there.

Answer 16

You are welcome. Glad to be able to help.