Prove cross-section of elliptic paraboloid is a ellipse

In summary, the question asks for the volume of a solid elliptic paraboloid with a horizontal cross-section at a certain height z being an ellipse. The equation for the ellipse is x^2/a^2+y^2/b^2<=(h-z)/h, 0<=z<=h. The solution involves rewriting the equation in standard ellipse form, calculating the area of the ellipse, and then integrating this area over the range of z. The resulting volume is (1/4)*h*π*b^2*(3-(b^2/a^2)).
  • #1
chris_usyd
39
0

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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  • #2
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:
 
  • #3
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?
 
  • #4
chris_usyd said:
… c is (0,h) …

what do you mean? :confused:

just write out the equation of the intersection
 
  • #5
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)??
 
  • #6
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 ?
 
  • #7
: )
then the equation is : hx^2/((h-z)*a^2) + hy^2/((h-z)*b^2)=1
this is an ellipse equation?
 
  • #8
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 ?
 
  • #9
(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
 

What is an elliptic paraboloid?

An elliptic paraboloid is a three-dimensional shape that resembles a bowl or a dome. It is formed by parabolas in two directions, creating a curved surface that opens up in one direction and is closed in the other direction.

What is a cross-section?

A cross-section is the two-dimensional shape that is created when a three-dimensional object is cut by a plane. It can be thought of as the "slice" of the object that is revealed by the cut.

How do you prove that the cross-section of an elliptic paraboloid is an ellipse?

To prove that the cross-section of an elliptic paraboloid is an ellipse, we can use the equation for an elliptic paraboloid in standard form: z = x^2/a^2 + y^2/b^2. By setting z to a constant value and solving for x and y, we can see that the resulting equation is that of an ellipse. This holds true for any constant value of z, thus proving that the cross-section of an elliptic paraboloid is always an ellipse.

Why is it important to prove the cross-section of an elliptic paraboloid is an ellipse?

Proving the cross-section of an elliptic paraboloid is an ellipse is important because it helps us understand the shape and properties of this three-dimensional figure. It also allows us to use our knowledge of ellipses to analyze and solve problems involving elliptic paraboloids.

Are there any real-world applications of this concept?

Yes, there are many real-world applications of this concept. Elliptic paraboloids can be found in architecture, engineering, and design. They are often used in the construction of domes, arches, and other curved structures. Understanding the cross-section of an elliptic paraboloid allows us to accurately design and build these structures with precise measurements and calculations.

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