Prove cross-section of elliptic paraboloid is a ellipse

[chris_usyd]

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?

 

[tiny-tim]

hi chris_usyd!

(try using the X² button just above the Reply box )

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 ?

 

[chris_usyd]

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?

 

[tiny-tim]

… c is (0,h) …

what do you mean?

just write out the equation of the intersection

 

[chris_usyd]

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)??

 

[tiny-tim]

but how to rewrite it as a standard ellipse form ( x^2/a^2+y^2/b^2=1)??

divide by (h-z)/h ?

 

[chris_usyd]

: )
then the equation is : hx^2/((h-z)*a^2) + hy^2/((h-z)*b^2)=1
this is an ellipse equation?

 

[chris_usyd]

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 ?

 

[tiny-tim]

(have a square-root: √ and a pi: π )

: )
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 ?

(difficult to read, but …) yes

 

[chris_usyd]

nice. ^^
thank you,T.

 

[chris_usyd]

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]

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()...

 

[tiny-tim]

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.)

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 ​

 

[chris_usyd]

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?

 

[tiny-tim]

sorry, I'm not following what you're doing …

you seem to be calculating ∫∫ zr drdθ ​

can you please explain in words how you're slicing the volume, and what you're integrating over?

 

[chris_usyd]

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

 

[chris_usyd]

: ((
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?

 

[chris_usyd]

oh about the r drdθ..
that is deltA=r drdθ..

 

[tiny-tim]

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

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)

 

[chris_usyd]

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!

 

[tiny-tim]

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 r²(θ₁ - θ₂)

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

sleep tight! :zzz:​

 

[TDR14]

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

 

[tiny-tim]

Welcome to PF!

Hi TDR14! Welcome to PF!

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

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

 

[TDR14]

Hi TDR14! Welcome to PF!


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

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

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?

 

[tiny-tim]

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 …

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

… 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.

 

[TDR14]

i don't understand …

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?

 

[tiny-tim]

you find the area A as a function of z, then the volume is ∫ A dz