underthesun
N1NJ4
- Joined
- Aug 10, 2002
- Messages
- 1,781
- Location
- At the top of Riovanes Castle
- Gender
- Undisclosed
- HSC
- 2010
Someone help me interpret it? my teacher left this as a homework question..
-
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
strange volumes Q (1 Viewer)
- Thread starter underthesun
- Start date
This is my interpretation of it: (sorry I can't draw a picture to accompany it)
The base of the solid is an ellipse (obviously), and when you slice this object perpendicular to the major segment (this means that each slice is parallel to the minor segment) the cross section of each slice will be in the shape of a parabola. This parabolic cross section always has a peak height of 4 units, hence the height of this 3D object is ALWAYS 4 units.
In case that explanation doesn't help, I'll try giving a mathematical description of the volume:
Let the ellipse described be:
(x^2)/2 + y^2 = 1 ... (1)
(so that the major axis lies on the x-axis)
Then each cross sectional parabolic section has an area A, which varies as x varies. To find A, you need to find the area of a certain section of a parabola.
Imagine a parabola that has two distinct real roots, the axis of this parabola is the y-axis (for simplicity), a maximum point at (0,4), and the distance between the roots of the parabola (along the x-axis) is equal to 2y (i.e. it has roots +-y)*. Find the area bounded by the parabola itself and the x-axis (by integration), and now you have A in terms of y (and you can make it in terms of x if you like from equation 1). Now that you have A in terms of x or y, integrate A over x or y with the appropriate limits, and you'll get the answer. I'll leave the calculation to you
.
* +- means plus or minus.
The base of the solid is an ellipse (obviously), and when you slice this object perpendicular to the major segment (this means that each slice is parallel to the minor segment) the cross section of each slice will be in the shape of a parabola. This parabolic cross section always has a peak height of 4 units, hence the height of this 3D object is ALWAYS 4 units.
In case that explanation doesn't help, I'll try giving a mathematical description of the volume:
Let the ellipse described be:
(x^2)/2 + y^2 = 1 ... (1)
(so that the major axis lies on the x-axis)
Then each cross sectional parabolic section has an area A, which varies as x varies. To find A, you need to find the area of a certain section of a parabola.
Imagine a parabola that has two distinct real roots, the axis of this parabola is the y-axis (for simplicity), a maximum point at (0,4), and the distance between the roots of the parabola (along the x-axis) is equal to 2y (i.e. it has roots +-y)*. Find the area bounded by the parabola itself and the x-axis (by integration), and now you have A in terms of y (and you can make it in terms of x if you like from equation 1). Now that you have A in terms of x or y, integrate A over x or y with the appropriate limits, and you'll get the answer. I'll leave the calculation to you
.* +- means plus or minus.
Last edited:
I understand it a little differently, I think that parabola is above the major axis
imagine an ellipse.
imagine a Parabola of peak height 4 units above the major axis.
then fit in the other parabolas.
underthesun
N1NJ4
- Joined
- Aug 10, 2002
- Messages
- 1,781
- Location
- At the top of Riovanes Castle
- Gender
- Undisclosed
- HSC
- 2010
I was thinking of the same graph affinity gave me, and I tried to work it out during the free period after the 4unit lesson.
But it seems to be something must be wrong, since the equation keeps giving 0 = 0 (strange, that's why). Here goes:
equation for ellipse:
[ :: (x^2)/4 + y^2 = 1 :: ]
then, the height for on the major axis is defined by
h = 4-(x^2)
got that equation by trying to figure k value etc..
When x = 0, then h = 4, which is the required condition. Further, when x = 2, or x = -2, h = 0, which is at the edge of the ellipse.
so the equation [ h = 4-(x^2) ] fits with the given condition.
Now, the problem is the cross-sections perpendicular to the major axis. Let's call this "uts" cross section..
then the height of the "uts" cross section is h, defined by the x position of the cross section relative to the ellipse.
and the width of the cross section is exactly the y value in terms of x, you see what I mean? i'll try post a pic later
then, the equation of this parabola is
u = h - k(y^2)
then u is the height of any "y" value.
when y = 0, u = h. This means, in the middle of the cross-section, which just happen to be in the major axis, the "altitude" is "h".
but when you try to find the k value of u, it seems impossible..
can you people understand what i'm rambling about? hope it's clear
But it seems to be something must be wrong, since the equation keeps giving 0 = 0 (strange, that's why). Here goes:
equation for ellipse:
[ :: (x^2)/4 + y^2 = 1 :: ]
then, the height for on the major axis is defined by
h = 4-(x^2)
got that equation by trying to figure k value etc..
When x = 0, then h = 4, which is the required condition. Further, when x = 2, or x = -2, h = 0, which is at the edge of the ellipse.
so the equation [ h = 4-(x^2) ] fits with the given condition.
Now, the problem is the cross-sections perpendicular to the major axis. Let's call this "uts" cross section..
then the height of the "uts" cross section is h, defined by the x position of the cross section relative to the ellipse.
and the width of the cross section is exactly the y value in terms of x, you see what I mean? i'll try post a pic later
then, the equation of this parabola is
u = h - k(y^2)
then u is the height of any "y" value.
when y = 0, u = h. This means, in the middle of the cross-section, which just happen to be in the major axis, the "altitude" is "h".
but when you try to find the k value of u, it seems impossible..
can you people understand what i'm rambling about? hope it's clear
underthesun
N1NJ4
- Joined
- Aug 10, 2002
- Messages
- 1,781
- Location
- At the top of Riovanes Castle
- Gender
- Undisclosed
- HSC
- 2010
affinity, what graph program do you use?
and can it actually calculate the volume, does it cost a lot of memory (RAM), take a lot of system resource, etc etc..
very interesting
and can it actually calculate the volume, does it cost a lot of memory (RAM), take a lot of system resource, etc etc..
very interesting
An interesting volume problem. Might just follow wogboy's excellent instruction.
Volume=int(2,-2){AdX}(trying to remember Spice Girl's integration notation) where A is the area of the elemental parabola of height 4, and sitting on ellipse at points (X,-Y) and (X,Y)
Family of parabolas with vertex (0,4) is
x^2=4a(y-4) the parabola we want cuts the x axis at +-Y.
Therefore a=-Y^2/16, and A=int(Y,-Y){(Y^2-x^2)4/Y^2 dx} ....
which gives A=(16/3)Y
Thus volume = int(2,-2){(16/3)YdX} , now
Y=sqrt(1-X^2/4) from ellipse x^2/4+Y^2/1=1
volume=int(2,-2){(16/3)sqrt(1-X^2/4)dX} a standard integral.
Note that the parabola at the extremities are just line segments of length 4 and gets "fatter" nearer the center.
Note also Affinity's volume model does not have a constant height of 4, and therefore must state how it behaves between max height of 4 and say min height of zero at extremities.
Volume=int(2,-2){AdX}(trying to remember Spice Girl's integration notation) where A is the area of the elemental parabola of height 4, and sitting on ellipse at points (X,-Y) and (X,Y)
Family of parabolas with vertex (0,4) is
x^2=4a(y-4) the parabola we want cuts the x axis at +-Y.
Therefore a=-Y^2/16, and A=int(Y,-Y){(Y^2-x^2)4/Y^2 dx} ....
which gives A=(16/3)Y
Thus volume = int(2,-2){(16/3)YdX} , now
Y=sqrt(1-X^2/4) from ellipse x^2/4+Y^2/1=1
volume=int(2,-2){(16/3)sqrt(1-X^2/4)dX} a standard integral.
Note that the parabola at the extremities are just line segments of length 4 and gets "fatter" nearer the center.
Note also Affinity's volume model does not have a constant height of 4, and therefore must state how it behaves between max height of 4 and say min height of zero at extremities.
I made that one with mathematica.
vague outline:
z = 4 - x^2 - 4y^2
integrate z W.R.T. y with limits + / - sqrt(1- x^2/4) then integrate the result with respect to x limits -2 to 2.
gets a little messy, but thats 4U
maybe k = 4?
vague outline:
z = 4 - x^2 - 4y^2
integrate z W.R.T. y with limits + / - sqrt(1- x^2/4) then integrate the result with respect to x limits -2 to 2.
gets a little messy, but thats 4U
it's parabolic, ie the parabola passing through points (-2,0,0), (2,0,0) and (0,0,4)
might as well put my solution up to my model
it turns out to be 4Pi Units cubed
Z= 4 - x<SUP>2</SUP>- 4y<SUP>2</SUP>
integrating z with respect to y with limits - (1/2)*sqrt(4 -x<SUP>2</SUP>)
and (1/2)*sqrt(4 - x<SUP>2</SUP>)
will give 2/3 (4 - x<SUP>2</SUP>)<SUP>3/2</SUP>
then integrating with respect to x with limits -2 , 2, using sub x= 2sin(t)
will give the volume = 4Pi
Last edited:
Hey,
nice question (volumes was my favourite 4U topic, it always seems like magic) and nice solution as well. A useful trick in questions with parabolas is to remember that Simpson's rule is exact for parabolas. So for your parabolas parallel to minor axis going through points (0,-0.5*sqrt(4-x^2),0), (0,0,4-x^2) and (0,0.5*sqrt(4-x^2),0).
A=(b-a)/6*[f(a) + 4*F((a+b)/2) + f(b)]
= sqrt(4-x^2)/6*[4(4-x^2)]
=2/3*(4-x^2)^3/2
as before.
I don't know if thats any quicker but it's pretty cool.
Martin
nice question (volumes was my favourite 4U topic, it always seems like magic) and nice solution as well. A useful trick in questions with parabolas is to remember that Simpson's rule is exact for parabolas. So for your parabolas parallel to minor axis going through points (0,-0.5*sqrt(4-x^2),0), (0,0,4-x^2) and (0,0.5*sqrt(4-x^2),0).
A=(b-a)/6*[f(a) + 4*F((a+b)/2) + f(b)]
= sqrt(4-x^2)/6*[4(4-x^2)]
=2/3*(4-x^2)^3/2
as before.
I don't know if thats any quicker but it's pretty cool.
Martin
underthesun
N1NJ4
- Joined
- Aug 10, 2002
- Messages
- 1,781
- Location
- At the top of Riovanes Castle
- Gender
- Undisclosed
- HSC
- 2010
"simpson's rule is exact for parabolas"
Are you sure? Can people quote that in tests and get full marks?
that sounds really good to me
Are you sure? Can people quote that in tests and get full marks?
that sounds really good to me
underthesun
N1NJ4
- Joined
- Aug 10, 2002
- Messages
- 1,781
- Location
- At the top of Riovanes Castle
- Gender
- Undisclosed
- HSC
- 2010
does simpson's rule only work to x<sup>2</sup>=4ay, or does that also work for any parabola y = 4 - x<sup>2</sup> etc..
Mathematician
Member
- Joined
- Nov 3, 2002
- Messages
- 188
...
in the 3u books why do they use the fundamental theorem of calculus to prove simpsons rule?
Use something that can give u exact areas and volumes to prove a formula that can give u an approximation???
Is it because using simpsons rule is faster and u can use it to find the area under any function even the really complex functions to integrate
???????????
in the 3u books why do they use the fundamental theorem of calculus to prove simpsons rule?
Use something that can give u exact areas and volumes to prove a formula that can give u an approximation???
Is it because using simpsons rule is faster and u can use it to find the area under any function even the really complex functions to integrate
???????????
Last edited: