- #1

Icebreaker

This has me stumped. The volume bounded by [tex]r^2=9-z[/tex] and [tex]z=0[/tex] is not closed in 3-space. But if they really meant region, triple-integrating a region with no volume gives 0. What should I do?

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- #1

Icebreaker

This has me stumped. The volume bounded by [tex]r^2=9-z[/tex] and [tex]z=0[/tex] is not closed in 3-space. But if they really meant region, triple-integrating a region with no volume gives 0. What should I do?

- #2

matt grime

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x^2 = 9-z and the x axis about the z axis. Why isn't that closed?

If we imagine the x-y plane on the floor with the positive z axis coming upwards, it's like integrating over the volume given by putting a cap on the floor.

- #3

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I tried to attach a graph,but the OS wouldn't let me...:yuck: Matt's right,it is a cap.

Daniel.

Daniel.

- #4

Icebreaker

Why would we rotate it? r^2 = 9 - z describes a parabolic cylinder.

- #5

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Nope,a revolution paraboloid around the "Oz" axis...

Daniel.

Daniel.

- #6

HallsofIvy

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All I can do is echo the others: z= 9- r^{2} is a paraboloid, with vertex at (0,0,9), opening downward, and "closing" it at z= 0 certainly does form a closed figure. When z= 0, r^{2}= 9 so r= 3. The integration (in polar coordinates) will be over [itex]0\le \theta\le 2\pi[/itex],[itex] 0\le r \le 3[/itex].

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- #7

Icebreaker

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Indeed, it should have been specified that cylindrical coordinates were being used.

- #9

Hippo

Icebreaker, use [tex] 9-r^2[/tex] as the upper limit in your integration of *f* over *z*.

[Edit: Wait, do you already get the idea? Sorry, if I'm being redundant.]

[Edit: Wait, do you already get the idea? Sorry, if I'm being redundant.]

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- #10

Icebreaker

- #11

HallsofIvy

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No, it is not a "trick" question and it **does** specify that cylindrical coordinates are to be used- your original post said "Compute the triple integral of f(r,θ,z), over the region bounded by the paraboloid and the plane"

What do you think f(r,θ,z) means?

What do you think f(r,θ,z) means?

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- #12

Hippo

- #13

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The way i know these variables,it should have been [itex]f\left(\rho,\varphi,z\right) [/itex].

Daniel.

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Why should it necessarily mean that they're using cylindrical coordinates? Just because the variables areWhat do you think f(r,θ,z) means?

Now, the fact that the question said that the region was bounded by a paraboloid might be grounds for a guess as to the parameterization...

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- #15

matt grime

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- #16

Icebreaker

f(r,θ,z) is simply a 3 variable function. There's no reason why I must treat it differently than f(a,b,c) or x(s,r,t). The point is that I must GUESS what the question is trying to ask, and that's the problem.HallsofIvy said:What do you think f(r,θ,z) means?

- #17

matt grime

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These are commonly accepted practices, even if some of us agree that it is not necessarily a good thing.

- #18

- #19

Hippo

The function was r^2.

Is that just a typo, or did you calculate the wrong integral?

Is that just a typo, or did you calculate the wrong integral?

- #20

matt grime

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An infinitesimal volume element is rdr(dtheta)dz in cylindrical polars.

- #21

Hippo

Oh, I remember that. :yuck:

- #22

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I'm not necessarily crtiticising the textbook at all. It's quite possible that it states at some point that it always refers to a certain parameterization when it uses certain symbols, which is perfectly fine with me, as long as it makes it clear that those symbols aren't parameterized that way *a priori*.

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