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Mariah's avatar

Can you help me understand what a parabolic cylinder is in the context of this problem?

Asked by Mariah (25785points) July 5th, 2011

I have the most ridiculous calc homework this week. I’m going crazy trying to figure out this problem:

Evaluate the triple integral ∫∫∫ E (x+y) dV where E is bounded by the parabolic cylinder y=5x^2 and the planes z=x, y=20x, and z=0.

I don’t want solutions, mostly what I want is some help understanding the parabolic cylinder. I’m very confused because y=5x^2 is just the equation of a parabola; it’s two dimensional and I don’t know how it relates to a cylinder in any way. Am I supposed to extrude that parabola parallel to the z axis? If so, can you give me some pointers on how I’m supposed to develop an equation for the resulting three-dimensional shape?

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6 Answers

WasCy's avatar

I’m just guessing here, but it sounds to me that instead of a normal cylinder with a circular wall, the cylinder you’re looking at here is ovoid (if closed), meaning that sections of it would have a parabolic curved wall.

Here are some examples, if it helps.

jaytkay's avatar

Could it be two parabolas with their feet together? Shaped kind of like a car muffler? I know that makes it not a function, though.

Or a parabola with one flat side?

Mariah's avatar

Okay, I don’t know wtf is wrong with me. I entered in 4 different answers to this problem yesterday, all of which the website said were incorrect.

So I did this problem over and over and over and over again and kept getting an answer that I (thought) had already been confirmed as incorrect.

Today I entered in that answer again, and it was right. I don’t know if I just made a typo the first time, or what. I’m dumb.

The way I approached the parabolic cylinder, which ended up getting me the right answer, was basically ignoring that is said it was a cylinder and just treating it as a parabola, oddly enough.

To solve an iterated triple integral, one method you can use is to take the two dimensional slice on the xy plane, use it to develop your x and y limits of integration, and then write z limits of integration in the form of z1(x,y) < z < z2(x,y). On the xy plane, x is bounded by 0 < x < 4, y is bounded below by 5x^2 and above by 20x, so those become the limits of integration on the two outermost integrals. z, I wasn’t sure if I could just say it was bounded by 0 and x, because if I was supposed to extrude the parabola, or something similar, that could put an upper bound on z below z = x, which I would have to use for z2(x,y). But, using 0 < z < x as the limits on the innermost integral yielded a correct answer. PHEW.

LostInParadise's avatar

It should have been called a circular paraboloid, which is what you get when you spin a parabola around the axis of symmetry.

gasman's avatar

Parabolic cylinder is correct. You have 3-d x-y-z coordinates. The boundary condition y=5x^2 is a parabola in the x-y plane at z=0. Since z does not appear in the equation it holds for all z, i.e., all planes parallel to the x-y plane. “Cylinder” here means the parabola is translated up & down in the z-direction, perpendicular to its plane, as in images linked by @WasCy. (A circle in the x-y plane instead of a parabola, would generate an ordinary cylinder parallel to the z axis.)

@Mariah You may have said it best: ”...extrude that parabola parallel to the z axis.”

@LostInParadise I’m not sure a surface of revolution is what is described.

LostInParadise's avatar

I stand corrected. Parabolic cylinder is what satisfies the equation.

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