Prove cross-section of elliptic paraboloid is a ellipse

AI Thread Summary
The discussion revolves around proving that the horizontal cross-section of an elliptic paraboloid at height z is an ellipse. The equation of the elliptic paraboloid is given as x²/a² + y²/b² ≤ (h-z)/h, where 0 ≤ z ≤ h. Participants explore how to rewrite this equation in standard ellipse form, concluding that dividing by (h-z)/h yields the standard form of an ellipse. They also discuss calculating the area of the ellipse at height z and the volume of the paraboloid using integration in polar coordinates. The conversation emphasizes the relationship between the parameters of the paraboloid and the resulting elliptical cross-sections.
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Homework Statement


a elliptic paraboloid is x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h, show that the horizontal cross-section at height z, is an ellipse


Homework Equations





The Attempt at a Solution


i don't know how to prove this, i only know that the standard ellipse is x^2/a^2+y^2/b^2=1..
could someone help me?
 
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hi chris_usyd! :smile:

(try using the X2 button just above the Reply box :wink:)
chris_usyd said:
a elliptic paraboloid is x2/a2+y2/b2<=(h-z)/h, 0<=z<=h, show that the horizontal cross-section at height z, is an ellipse

what is the intersection of it with the plane z = C ? :wink:
 
yes Tim, i think the intersection is the plane z=C, and c is (0,h);
but how to rewrite this to standard ellipse form?
 
chris_usyd said:
… c is (0,h) …

what do you mean? :confused:

just write out the equation of the intersection
 
z is between 0 and h...sorry my fault.

i think the equation of the intersection @certain height 'z' is just x^2/a^2+y^2/b^2<=(h-z)/h..isn't it?
but how to rewrite it as a standard ellipse form ( x^2/a^2+y^2/b^2=1)??
 
chris_usyd said:
but how to rewrite it as a standard ellipse form ( x^2/a^2+y^2/b^2=1)??

divide by (h-z)/h ?
 
: )
then the equation is : hx^2/((h-z)*a^2) + hy^2/((h-z)*b^2)=1
this is an ellipse equation?
 
if so, the area of the intersection @certain height is pi*((h-z)/h)*a*b, isn't it?

since @ height z, when x=0, y=sqrt((h-z)/h)*b,when y=0, x=sqrt((h-z)/h)*a.
to calculate the area, just simply pi*sqrt((h-z)/h)*sqrt((h-z)/a),which is pi*((h-z)/h)*a*b ?
 
(have a square-root: √ and a pi: π :wink:)
chris_usyd said:
: )
then the equation is : hx^2/((h-z)*a^2) + hy^2/((h-z)*b^2)=1
this is an ellipse equation?
chris_usyd said:
if so, the area of the intersection @certain height is pi*((h-z)/h)*a*b, isn't it?

since @ height z, when x=0, y=sqrt((h-z)/h)*b,when y=0, x=sqrt((h-z)/h)*a.
to calculate the area, just simply pi*sqrt((h-z)/h)*sqrt((h-z)/a),which is pi*((h-z)/h)*a*b ?

(difficult to read, but …) yes :smile:
 
  • #10
nice. ^^
thank you,T.
 
  • #11
Tim, same equation for the solid elliptic paraboloid, x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h.
the question is : suppose a>=b, calculate the volume of that part of the paraboloid that lies above the disc x^2+y^2<=b^2.( use a suitable integral in polar coordinates.)
 
  • #12
how can i set up x and y in polar coordinates??
we usually get a cylinder or whatever the intersection is a circle.
in these cases, x=rcos(),y=rsin()...
 
  • #13
chris_usyd said:
Tim, same equation for the solid elliptic paraboloid, x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h.
the question is : suppose a>=b, calculate the volume of that part of the paraboloid that lies above the disc x^2+y^2<=b^2.( use a suitable integral in polar coordinates.)

chris_usyd said:
how can i set up x and y in polar coordinates??
we usually get a cylinder or whatever the intersection is a circle.
in these cases, x=rcos(),y=rsin()...

let's see …

the paraboloid sits on a base on the x-y plane which is the ellipse x²/a²+y²/b² = 1,

and the ellipses get smaller up to height z = h, where they disappear

and we want the volume of that inside the cylinder that just fits inside the base ellipse​

ok, so slice it into horizontal "discs" of thickness dh at height h …

each "disc" is actually the intersection of a circle (r = b) with an ellipse

that's the area which you need the polar coordinates to find :wink:
 
  • #14
ok... i do this question like this.. tell me if i am right or not. : ))
let x = r∙cos(θ),y = r∙sin(θ)
dS = dxdy = r drdθ
because
x² + y² ≤ b² =>r²∙cos²(θ) + r²∙sin²(θ) ≤ b²=>r² ≤ b²
and r is the distance to the origin, which can not be negative. So the range of integration in radial direction is:
r:0->b
θ:0->2pi
the height above the disc is z = h∙(1 - (x²/a²) - y²/b²)
=> h∙(1 - (1/a²)∙r²∙cos²(θ) - (1/b²)∙r²∙sin²(θ))

with the identities
cos(2∙θ) = cos²(θ) - sin²(θ) = 2∙cos²(θ) - 1 = 1 - 2∙sin²(θ)
=>
cos²(θ) = (1/2)∙(cos(2∙θ) + 1)
sin²(θ) = (1/2)∙(1 - cos(2∙θ))
height z can be rewritten as:
z = h∙(1 - (1/a²)∙r²∙(1/2)∙(cos(2∙θ) + 1) - (1/b²)∙r²∙(1/2)∙(1 - cos(2∙θ)))
= (1/2)∙h∙(2 - ((1/a²) + (1/b²))∙r² - ((1/a²) - (1/b²))∙r²∙cos(2∙θ))

Hence,
... b.. 2∙π
V = ∫... ∫ (1/2)∙h∙(2 - ((1/a²) + (1/b²))∙r² - ((1/a²) - (1/b²))∙r²∙cos(2∙θ))∙r dθdr
... 0.. 0
.. b
= ∫ (1/2)∙h∙( [2 - ((1/a²) + (1/b²))∙r²]∙(2∙π - 0)- (1/2)∙((1/a²) - (1/b²))∙r²∙(sin(4∙π) - sin(0))∙r dr
. 0
.. b
= ∫ (1/2)∙h∙(2 - ((1/a²) + (1/b²))∙r²)∙2∙π∙r dr
. 0
.. b
= ∫ h∙π∙( 2∙r - ((1/a²) + (1/b²))∙r³ dr
. 0
= h∙π∙{ (b² - 0²) - (1/4)∙((1/a²) + (1/b²))∙(b⁴ - o⁴) }
= h∙π∙( b² - (1/4)∙(b⁴/a²) - (1/4)∙b²
= (1/4)∙h∙π∙b²∙(3 - (b²/a²))
?
right?
 
  • #15
sorry, I'm not following what you're doing :redface:

you seem to be calculating ∫∫ zr drdθ :confused:

can you please explain in words how you're slicing the volume, and what you're integrating over?
 
  • #16
OK.Tim, this is what i learn from my textbook.
If f(x,y)>=0 for all (x,y) in some region R, then z= f(x,y) represents a surface sitting above the xy-plane and over R. the double integral f(x,y) dxdy can then be interpreted as the colume of the solid under the surface z=f(x,y),over R, since the volume of a small strip sitting over deltA=delt(x)*delt(y) is approximately f(x,y)*deltA
 
  • #17
: ((
i am confused as well, but this is what is written in my textbook-'course notes for Math2061", University of Sydney, school of mathematics and statistics.
basically i think it is just using double integral to find the volume...
Tim, does that make sense?
 
  • #18
oh about the r drdθ..
that is deltA=r drdθ..
 
  • #19
chris_usyd said:
If f(x,y)>=0 for all (x,y) in some region R, then z= f(x,y) represents a surface sitting above the xy-plane and over R. the double integral f(x,y) dxdy can then be interpreted as the colume of the solid under the surface z=f(x,y),over R, since the volume of a small strip sitting over deltA=delt(x)*delt(y) is approximately f(x,y)*deltA

ah, now i see :smile:

yes, that is a correct way of finding the volume …

you slice the volume into square vertical columns of height z = f(x,y) and base dxdy (and therefore volume x dxdy), and then sum all the individual volumes

so the volume is ∫∫ f(x,y) dxdy, = ∫∫ f(x,y)r drdθ​

but it's not the only way … you can slice the volume other ways, which may be easier

in this case, since the previous parts of the question are all about the horizontal cross-sections,

i'm guessing that they intended you to use horizontal slices, find the area of the intersection of that circle-and-ellipse (the circle part is easy, the ellipse part fairly easy), and then integrate that area over z

(your solution looks ok, but i haven't checked right through it)
 
  • #20
thanks Tim, in part a) of this question, i found the semi-axes of this elliptic paraboloid are b*sqrt((h-z)/h) and a*sqrt((h-z)/h)
therefore the intersection of the ellipse at height z is going to be [a*sqrt((h-z)/h)]*[b*sqrt((h-z)/h) ]*pi, isn't? which is just pi*(h-z)/h*a*b.
then how am i going to use polar coordinates? because a and b are constant numbers...

: )) tim its 11pm in sydney now, time for bed...
if u reply my post, i might not be able to read it tonight, but i will check tomorrow moring.. thanks for your help, have a good day!
 
  • #21
chris_usyd said:
then how am i going to use polar coordinates? because a and b are constant numbers...

you need to find the four values of θ where the circle and ellipse meet

over two sections, that's just the area of a sector of a circle, 1/2 r21 - θ2)

over the other two sections, it's the area of a sector of an ellipse, which is … ? :wink:

sleep tight! :zzz:​
 
  • #22
chris_usyd said:
if so, the area of the intersection @certain height is pi*((h-z)/h)*a*b, isn't it?

since @ height z, when x=0, y=sqrt((h-z)/h)*b,when y=0, x=sqrt((h-z)/h)*a.
to calculate the area, just simply pi*sqrt((h-z)/h)*sqrt((h-z)/a),which is pi*((h-z)/h)*a*b ?

I'm confused about how you got the area?

PS I'm doing this same course and assignment too
 
  • #23
Welcome to PF!

Hi TDR14! Welcome to PF! :smile:
TDR14 said:
I'm confused about how you got the area?

There's various ways of doing it, but the simplest is to note that an ellipse is a squashed circle, by a factor b/a along the minor axis,

so the area is πa2*b/a, = πab :wink:

(technically, one proves this with the substitution y' = ay/b o:))
 
  • #24


tiny-tim said:
Hi TDR14! Welcome to PF! :smile:


There's various ways of doing it, but the simplest is to note that an ellipse is a squashed circle, by a factor b/a along the minor axis,

so the area is πa2*b/a, = πab :wink:

(technically, one proves this with the substitution y' = ay/b o:))


ok, I know that.

But how do you put it in terms of x,y,z etc?

I got A=∏*(x/sqrt((h-z)/h))*(y/sqrt((h-z)/h))

Then I just follow chris_usyd's way of integrating and so on and so forth?
 
  • #25
TDR14 said:
But how do you put it in terms of x,y,z etc?

I got A=∏*(x/sqrt((h-z)/h))*(y/sqrt((h-z)/h))

i don't understand :confused:

A does not depend on x or y, only on z, see chris's formula …
chris_usyd said:
… therefore the intersection of the ellipse at height z is going to be [a*sqrt((h-z)/h)]*[b*sqrt((h-z)/h) ]*pi, isn't? which is just pi*(h-z)/h*a*b.
 
  • #26
tiny-tim said:
i don't understand :confused:

A does not depend on x or y, only on z, see chris's formula …

ok I get it now, then how do you find volume by slicing?
 
  • #27
you find the area A as a function of z, then the volume is ∫ A dz
 
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