Representing three-dimensional shapes with functions and integrating

• xwolfhunter
In summary, the conversation involves the speaker expressing their dislike for calculus class but their interest in math. They mention trying to solve a problem involving integrating a three-dimensional object using functions. They ask for clarification on their method and mention wanting to learn more about multiple integrals and differential equations. They also mention their lack of interest in physics and their confidence in understanding single-variable calculus.
xwolfhunter
So, I don't like calculus class, since it's very boring, but I do like math, and I intend to sort of become a mathematical autodidact. So I just thought I'd try to come up with a solution to a problem I created, and this was to integrate one of those gigantor Reese's easter egg things (holy cow, they're huge). Now, since it's just high school calculus, we don't do three-dimensional stuff, so I wanted to know how to represent a three-dimensional object using functions. After some thought, I figured that it was done simply by presenting each of the three planes with a function, and then where they intersect is your shape. Firstly, is this correct? Pretty sure it is, but I don't really know.
Secondly, I wanted to integrate a three-dimensional object that was not a two-dimensional function spun around its axis. So I thought up two functions,

$y=x^2$ if $y≤4$
$z=-y^2+16$

because they work well together, and the shape they produced (according to how I thought they'd do it) was good.

So, the way I thought I'd determine the volume was this,
$$V=\int_0^4 (-y^2+16)\,dy*\int_0^2 x^2\,dx$$
which results in $17\frac{2}{3}$.

Then I tried to split the volume into four equal parts along the z-axis, but I messed up somewhere (though if I'm right about the volume I'm right about my method, I know this) and haven't finished the calculations.

So if you could let me know if I'm doing this right, I'd appreciate it, and if I'm doing it wrong, please tell me how to do it right.

Thanks!

xwolfhunter said:
After some thought, I figured that it was done simply by presenting each of the three planes with a function, and then where they intersect is your shape. Firstly, is this correct? Pretty sure it is, but I don't really know.

I don't see what you are getting at here, but the way it is stated, this is not correct. The intersection of any number of planes will be at most 2 dimensional.

I don't see how your functions would represent a Reese's Easter Egg. Either way, it isn't quite the method you want. You should look up multiple integrals if this is something you want to learn. Most likely the function you want to integrate is something along the lines of ##z=5-\left(\frac12x^2+y^2\right)##, but you can't just multiply the integrals together (well, you can, but it does not mean what you want it to mean). Try graphing that and see if it is what you want. Then look up multiple integrals. Check back if you have questions.

If you are feeling bored with your class, I wouldn't jump to completely new material. Either work some harder problems in single variable calculus, or get into the nitty gritty. I don't mean to discourage you from this particular problem; by all means, follow it through to the end. It is my opinion that you would be better served by finding out what you haven't learned in single variable calculus first.

Are you in an AP class? If so, you are probably talking about differential equations now (or about to be) That is very important.
If you are also taking a physics class, use what you have learned in calculus and transfer it to integrating charge densities or center of mass or moment of inertia.
If you want the pure math side of things, go back and really dive into the definitions of continuity and differentiability.

Apologies for my ambiguity in my wording. Yes, I'm in AP Calc and AP Physics, but to be honest I am really no longer interested in the slightest in the physics class due to a problem involving simple harmonic motion wherein we assumed no friction on the ground, but a loss of energy due to friction in a collision (it was just irksome to the extreme), and pure mathematics is in my opinion more interesting (though I am intrigued by theoretical physics on the quantum level and the physics of space, the building blocks for which, I know, are the concepts I'm working on now in class) than fluid and thermodynamics.

The integrals were not intended to be representative of the easter egg, and I may have technically misspoken when I said intersection. I meant the area encapsulated when the two-dimensional functions, as is their wont, extend along the unspecified third dimension and "intersect". The planes on which the functions exist intersect and form the y-axis. Because y=3 and x=4 form a rectangle the area of which can be determined by multiplying their "integrals" from 0 to 3 and 0 to 4 respectively, I assumed, and I think honestly rightly, that this would apply to three-dimensional shapes defined by two-dimensional boundaries. I also determined that if two two-dimensional areas are multiplied together, regardless of their encapsulating shape, the result would be a given number, and no number other than that, so I disregarded the possibility of having to work with the wonky-shaped boundaries formed by my functions. Hence, I found the area of one of the encapsulating sides, and multiplied it by the area of the other.

Either I am so horribly misguided by mathematics that I am missing an extremely obvious blunder, or I simply stated the problem so confusingly that it was misinterpreted, for which, if this is the case, I apologize.

And I graphed that function, and no, it's not anything like that. There is a parabola on the x,y axis and a parabola on the y,z axis. I integrated their areas and then multiplied them together.

As for differential equations, I was never given a technical definition on them, so excuse me while I look that up on wikipedia. :)

Ah. No, we have not done differential equations. I will look into those.
As for calculus, beyond differential equations and anything else I do not know about, single-variable calculus is mostly in my grasp. The only issues might be obscure algebraic ones, since I never really payed attention in any of my algebra classes, but beyond that I can differentiate and integrate out the wazoo, since it's practically the end of the school year. The only thing that I want to do with single-variable calculus is find out how to use it in useful mathematical situations, the solutions of which might help me to step to the next layer of mathematics.

By the way, I'm assuming that you've progresses beyond calculus. I plan on teaching myself multi-variable calculus over the summer - once I've done that, what should I do next? What's after calculus?

As for the definitions of continuity and differentiability, unless there's some field of mathematics I'm unaware of which delves into great detail on those subjects, we dealt with those months ago. To the best of my knowledge, I know those things inside out and backwards to the fullest extent which they provide. At least conceptually.

And I don't know how much I could be discouraged from my excitement about math by somebody on a forum. Would you be? :)

After rereading, maybe some things sound abrasive, I can't really tell anymore, but if so, it was for the sake of brevity, not anything else, I assure you. That is why I'm not changing any of it.

xwolfhunter said:
The integrals were not intended to be representative of the easter egg, and I may have technically misspoken when I said intersection. I meant the area encapsulated when the two-dimensional functions, as is their wont, extend along the unspecified third dimension and "intersect". The planes on which the functions exist intersect and form the y-axis. Because y=3 and x=4 form a rectangle the area of which can be determined by multiplying their "integrals" from 0 to 3 and 0 to 4 respectively, I assumed, and I think honestly rightly, that this would apply to three-dimensional shapes defined by two-dimensional boundaries. I also determined that if two two-dimensional areas are multiplied together, regardless of their encapsulating shape, the result would be a given number, and no number other than that, so I disregarded the possibility of having to work with the wonky-shaped boundaries formed by my functions. Hence, I found the area of one of the encapsulating sides, and multiplied it by the area of the other.

Multiplying the area in one plane by the area in another will give you a 4D space. If you want a 3D space you will have to use a double integral. Beyond that, I'm not sure I follow any of your logic.

As for differential equations, I was never given a technical definition on them, so excuse me while I look that up on wikipedia. :)

This is part of the AP curriculum. Only the most basic differential equations, but the idea is very important.

By the way, I'm assuming that you've progresses beyond calculus. I plan on teaching myself multi-variable calculus over the summer - once I've done that, what should I do next? What's after calculus?

I would first study the normal single variable topics. Have you covered integration by parts, by partial fractions, Taylor series, calculus in polar coordinates, and two dimensional parametric equations? If you have covered all of the AP BC curriculum, then multivariable would be a good step. Certainly you could skip to multivariable calc, but there is a lot of single variable calculus beyond what is covered in an AP AB class (which most are).

As for the definitions of continuity and differentiability, unless there's some field of mathematics I'm unaware of which delves into great detail on those subjects, we dealt with those months ago. To the best of my knowledge, I know those things inside out and backwards to the fullest extent which they provide. At least conceptually.

Topology. Even without going into the generality of topology, there is a lot to think about with continuity that you have not considered. Sure, conceptually it is easy, but math is about the details. If you want an idea, check out baby Rudin (you can find it without buying the copy on amazon). It isn't the easiest intro though. I wouldn't really consider calculus at this level until after some multivariable calculus. Usually this level of rigor isn't attempted until after the normal three semesters of calculus.

We've just completed the AP AB calc curriculum. Very exciting. And yeah, the differential section was very sparse.
And I have a calculus book in which is found all of those concepts, which I'm reading. Started Taylor series'. Are they similar to Fourier series'? (I think that's what it is, where you form trig functions of graphs which are not functions, due to vertical segments, right?) Approximating a function with a converging series, or something.

And holy mackerel, you're right! I never though of that. Four dimensional shape . . . took a little thought, but that makes complete sense. So conceptually, I'd have to multiply the length of one of the infinitesimals by the length of the corresponding one on the other function . . . and then . . . the length of the . . . third dimension . . .agh my brain.

I'll look into double integrals. It'd take a better mind than mine to figure out how to do it without outside help. :) Though I desperately want to try, I just don't have a depth of knowledge in mathematics. Like I still don't know why the $dx$ is necessary in an integral. It's actually kinda silly that we don't go into the actual notation - I had to look up on my own to find that $d$ doesn't mean derivative, it means "an infinitesimally small bit of." Even knowing that, I still have no idea what $\frac{d}{dx}$ actually means, even though it's the same as $\frac{dy}{dx}$, which I understand conceptually. Nor can I see how $d[f(x)]$ is equivalent to $f'(x)$. You know what I'm saying? I know they all mean derivative, but I don't know what each one represents conceptually. And without that, all I know is how to do the stuff, not why, which is useless when it comes to actual thought instead of just solving homework problems.

1. How are three-dimensional shapes represented with functions?

In mathematics, three-dimensional shapes can be represented using functions such as parametric equations or implicit equations. These equations describe the relationship between three coordinates (x, y, z) and can be graphed to create a visual representation of the shape.

2. What is the purpose of integrating when representing three-dimensional shapes?

Integrating is used to find the volume and surface area of three-dimensional shapes. By using integration, we can calculate the area under a curve or the volume of a solid, which are important properties of three-dimensional shapes.

3. Can any three-dimensional shape be represented with a function?

No, not all three-dimensional shapes can be represented with a function. Some shapes, such as fractals or irregular shapes, may require more complex equations or a combination of functions to accurately represent them.

4. How do you integrate a function to find the volume of a three-dimensional shape?

To find the volume of a three-dimensional shape, we use a method called triple integration. This involves integrating the function with respect to x, y, and z, within the boundaries of the shape. The resulting value will be the volume of the shape.

5. Are there any real-world applications of representing three-dimensional shapes with functions and integrating?

Yes, there are many real-world applications of using functions and integration to represent three-dimensional shapes. For example, in engineering and architecture, these techniques are used to design and analyze structures, while in physics and chemistry, they are used to model and understand complex systems and molecules.

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