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Prove crosssection of elliptic paraboloid is a ellipse 
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#19
May2212, 07:30 AM

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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 drdθbut 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 crosssections, i'm guessing that they intended you to use horizontal slices, find the area of the intersection of that circleandellipse (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
May2212, 07:43 AM

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thanks Tim, in part a) of this question, i found the semiaxes of this elliptic paraboloid are b*sqrt((hz)/h) and a*sqrt((hz)/h)
therefore the intersection of the ellipse at height z is gonna be [a*sqrt((hz)/h)]*[b*sqrt((hz)/h) ]*pi, isn't? which is just pi*(hz)/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!! 


#21
May2212, 07:55 AM

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over two sections, that's just the area of a sector of a circle, 1/2 r^{2}(θ_{1}  θ_{2}) over the other two sections, it's the area of a sector of an ellipse, which is … ? sleep tight! 


#22
May2212, 09:10 AM

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PS I'm doing this same course and assignment too 


#23
May2212, 12:27 PM

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Hi TDR14! Welcome to PF!
so the area is πa^{2}*b/a, = πab (technically, one proves this with the substitution y' = ay/b ) 


#24
May2212, 08:09 PM

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But how do you put it in terms of x,y,z etc? I got A=∏*(x/sqrt((hz)/h))*(y/sqrt((hz)/h)) Then I just follow chris_usyd's way of integrating and so on and so forth??? 


#25
May2312, 04:04 AM

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


#26
May2312, 01:18 PM

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