# Prove cross-section of elliptic paraboloid is a ellipse

by chris_usyd
Tags: ellipse, elliptic paraboloid
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P: 26,148
 Quote by chris_usyd 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)
 P: 39 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!!
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P: 26,148
 Quote by chris_usyd then how am i gonna 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 … ?
sleep tight!
P: 3
 Quote by 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 ????
I'm confused about how you got the area?

PS I'm doing this same course and assignment too
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P: 26,148
Hi TDR14! Welcome to PF!
 Quote by TDR14 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 )
P: 3
 Quote by tiny-tim 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???
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P: 26,148
 Quote by TDR14 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 …
 Quote by chris_usyd … 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.
P: 3
 Quote by tiny-tim 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?
 Sci Advisor HW Helper Thanks P: 26,148 you find the area A as a function of z, then the volume is ∫ A dz

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