# Prove cross-section of elliptic paraboloid is a ellipse

1. May 19, 2012

### chris_usyd

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. The attempt at a solution
i dont 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?

2. May 20, 2012

### tiny-tim

hi chris_usyd!

(try using the X2 button just above the Reply box )
what is the intersection of it with the plane z = C ?

3. May 20, 2012

### 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?

4. May 20, 2012

### tiny-tim

what do you mean?

just write out the equation of the intersection

5. May 20, 2012

### 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)??

6. May 20, 2012

### tiny-tim

divide by (h-z)/h ?

7. May 20, 2012

### chris_usyd

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

8. May 20, 2012

### chris_usyd

if so, the area of the intersection @certain height is pi*((h-z)/h)*a*b, isnt 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. May 20, 2012

### tiny-tim

(have a square-root: √ and a pi: π )
(difficult to read, but …) yes

10. May 20, 2012

### chris_usyd

nice. ^^
thank you,T.

11. May 22, 2012

### 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.)

12. May 22, 2012

### 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()...

13. May 22, 2012

### tiny-tim

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

14. May 22, 2012

### 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???

15. May 22, 2012

### 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?

16. May 22, 2012

### 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

17. May 22, 2012

### 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?

18. May 22, 2012

### chris_usyd

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

19. May 22, 2012

### tiny-tim

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)

20. May 22, 2012

### 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 gonna 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 gonna 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!!