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## Homework Statement

Not sure how to approach this one, does anyone have an idea?

BTW, in the top line it should read xz-plane, not xy.

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- Thread starter diffusion
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Not sure how to approach this one, does anyone have an idea?

BTW, in the top line it should read xz-plane, not xy.

- #2

tiny-tim

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(try using the X

Forget all the stuff about the cylinder …

the question is simply to prove that if you rotate z = e

that's easy, isn't it?

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(try using the X^{2}tag just above the Reply box )

Forget all the stuff about the cylinder …

the question is simply to prove that if you rotate z = e^{-x2}about the z-axis, you get z = e^{-x2-y2}…

that's easy, isn't it?

I'm sure it's easy once I've been set in the right direction. I can

- #4

tiny-tim

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Well you will rotate the function around the z-axis, so that z = f(x) becomes z = f(x,y). Not sure how to express it as an equation, besides telling you that the equation is a function of both x and y.

- #6

tiny-tim

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Try putting it in words first …

what happens to each individual point on the original curve?

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Try putting it in words first …

what happens to each individual point on the original curve?

Produces a circle when rotated.

- #8

tiny-tim

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Produces a circle when rotated.

Yup! And the equation of a circle is … ?

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Yup! And the equation of a circle is … ?

Yep, I recognized this before I posted the question. Again I just don't know how to

"Since each point on the graph z = e

I don't know, somehow this justification seems vague and inadequate.

- #10

tiny-tim

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"Since each point on the graph z = e^{-r2}will generate a circle when rotated about the z-axis, and the equation for a circle is r^{2}= x^{2}+ y^{2}, we can make this substitution into our equation for r^{2}, giving us z = e^{-(x2+y2)}."

ok, let's formalise that …

the point (x

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ok, let's formalise that …

the point (x_{0},0,z_{0}) generates the circle (x,y,z), where … ?

At z

- #12

tiny-tim

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