Engage your mind in a Math Problem

Discus: SAT/ACT Tests and Test Preparation: July 2004 Archive: Engage your mind in a Math Problem
 By Michael_Pham (Michael_Pham) on Wednesday, July 07, 2004 - 09:09 pm: Edit

The problem is:

There is a cube with length 1' and a height of 2'.
Inscribed in the cube is a 3D Parabola with vertex O in the center at the bottom of the cube and the 3D parabola touches each side of the cube.

Find the total area of the space inside the cube that is not taken up by the 3D parabola.

Find the area of each individual section in the cube not taken up by the 3D parabola.

Maybe the term 3D parabola isn't correct, but it is not a cone. It is a parabola with a circle that closes it up at the top.

I could probably solve the first part of the problem if I knew the formula for the area of the 3D parabola.

 By Joshjmgs (Joshjmgs) on Wednesday, July 07, 2004 - 09:24 pm: Edit

There is a cube with length 1' and a height of 2'.

Thats not a cube.

 By Michael_Pham (Michael_Pham) on Wednesday, July 07, 2004 - 09:46 pm: Edit

Oops.. rectangular prism.

1 x 1 x 2

 By Al0 (Al0) on Wednesday, July 07, 2004 - 09:53 pm: Edit

A cube can't have a length of 1 and height of 2; by definition,a cube is that it has all equal sides.

the 3D parabola you're thinking of is probably a paraboloid (a free rotation of a parabola about its axis of symmetry) whose general equation is x^2/a^2+y^2/b^2=z where a and b are constants.

To solve your problem one whould need to know the width of this so-called "cube," since apparently its dimensions are not all the same. From there, you would define a parabolic cross section of the paraboloid in the first and second quadrants and use the shell or disk method (calculus) to calculate the volume of the free rotation of said parabola. Since the paraboloid is inscribed inside a rectangular prism, there would only be one region inside the cube not occupied by the paraboloid so there would not be any individual sections.

Define the vertex of a parabola at (0,0); one point at (-w/2,h) and another at (w/2,h) with w being the width and h being the height.

Substitute these coordinate pairs into the general from of a quadratic (y=ax^2+bx+c) and get:
c=0
h=aw^2/4-bw/2
h=aw^2/4+bw/2
use linear combination on the second and third equations:
2h=aw^2/2
4h=aw^2
a=4h/w^2
substituting this into the second of the simultaneous equations:
h=h-bw/2
0=-bw/2
b=0
So the general from of a parabola with width w and height h and vertex at (0,0) is y=(4hx^2)/(w^2).
Using disc method:
x=sqrt(yw^2/(4h))
Volume=pi*fnInt(r(y)^2,y,0,h)
V=pi*fnInt(yw^2/(4h),y,0,h)
V=pi*w^2/(4h)*y^2/2]0,h
V=pi*w^2/(4h)*(h^2/2)
V=pi*h*w^2/8
or shell method:
Volume=2*pi*fnInt(r(x)*h(x),x,0,w/2)
V=2*pi*fnInt((h-4hx^2/w^2)*x,x,0,w/2)
V=2*pi*fnInt(hx-4hx^3/w^2,x,0,w/2)
V=2*pi*(hx^2/2-hx^4/w^2]0,w/2
V=2*pi*(hw^2/8-hw^4/(16w^2)-0)
V=2*pi*(hw^2/8-hw^2/16)
V=2*pi*(hw^2/16)
V=pi*h*w^2/8

So the volume of a paraboloid with height h and width w is pi*h*w^2/8.

 By Al0 (Al0) on Wednesday, July 07, 2004 - 09:57 pm: Edit

So with height two and width one, the volume of the paraboloid is pi*2*1^2/8=pi/4~0.785 and the area outside of the paraboloid, but inside the cube is 2x1x1-0.785=1.215 and like I said before, I think that there is only one region inside the cube and outside the paraboloid.

 By Michael_Pham (Michael_Pham) on Wednesday, July 07, 2004 - 11:20 pm: Edit

Wow thanks Alo!!

You're good at this type of math.

How can I better improve my abstract math skills?

I can often do regular problems with ease, but when it comes to problems with these shapes, I have a hard time.
I've just finished Alg2./Trig. (the teacher didn't reach trig)
and I'm going to pre-calc. next year

 By Al0 (Al0) on Thursday, July 08, 2004 - 12:05 am: Edit

I'm going to be a senior/freshman (my parents don't want to pay for another year of high school, but it is kind of late to go to a good college). Just work on your visualization skills and take anything your teachers teach you ten steps further. Like next year some good pre-calc things are doing geometric proofs of some of the limit identities you get (limx-> sinx/x=1,etc.) or doing some of the summation formulas (just scratching out things in the margins of my other notebooks I got the identities up to i^12 by hand). Also you can just try and describe objects mathematically (via equations to be able to see in 3D space the result of an equation). Basically, don't just be complacent and do only the math your teachers tell you to (maybe I just say that because I have been less than enamored of some of mine).