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Finding volume bounded by paraboloid and cylinder

by iqjump123
Tags: cylinder, integral, paraboloid, volume
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iqjump123
#1
Jul1-11, 10:13 AM
P: 57
1. The problem statement, all variables and given/known data
Find the volume bounded by the paraboloid z= 2x2+y2 and the cylinder z=4-y2. Diagram is included that shows the shapes overlaying one another, with coordinates at intersections. (Will be given if necessary)


2. Relevant equations
double integral? function1-function2?


3. The attempt at a solution
I saw from previous threads involving volumes, but still am lost when I try to do my own problem :\ Most paraboloid involving problems start by changing to polar coordinates- should I do it for this one? I know that at the end it will end up being a double integral, but I am not sure how to set it up.

physics forums have been a big help. Thanks!
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vela
#2
Jul1-11, 12:54 PM
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Are your equations correct because z=4-y2 isn't the equation of a cylinder?
HallsofIvy
#3
Jul1-11, 03:53 PM
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Quote Quote by vela View Post
Are your equations correct because z=4-y2 isn't the equation of a cylinder?
Yes, it is. z= 4- y2 is a parabola in the yz-plane and, extended infinitely in the x-direction, is a parabolic cylinder, though not, of course, a circular cylinder.

vela
#4
Jul1-11, 04:56 PM
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Finding volume bounded by paraboloid and cylinder

D'oh!
iqjump123
#5
Jul4-11, 11:09 PM
P: 57
thanks for the reply- yes, just like what hallsofivy mentioned, the equations are correct.
At this point, I am still lost, however. Any other suggestions? Thanks!
vela
#6
Jul5-11, 12:06 AM
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As you mentioned in your original post, you want to calculate something like
[tex]V = \iint\limits_A [z_1(x,y)-z_2(x,y)]\,dy\,dx[/tex]
iqjump123
#7
Jul6-11, 11:39 PM
P: 57
Hello vela,
thanks for the reply!
Link below is an image of the problem image that was given:


Uploaded with ImageShack.us

I figured that the bounds of dy will stretch from 0 to sqrt(2-x^2), and dx will stretch from 0 to sqrt(2). Is this correct? Thanks!
vela
#8
Jul7-11, 12:48 AM
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Why are the lower limits 0 for both x and y?
iqjump123
#9
Jul7-11, 07:21 AM
P: 57
Quote Quote by vela View Post
Why are the lower limits 0 for both x and y?
Well I assumed so, since the problem shape starts from the 0 position for all 3 coordinate systems. Is this approach not correct?
vela
#10
Jul7-11, 11:34 AM
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Does the original problem statement say the solid is bounded by the x=0, y=0, and z=0 planes or something equivalent? If it does, your limits look fine. I know the picture suggests this, but you never mentioned it in the original post, nor does it appear in your scan.
iqjump123
#11
Jan15-12, 08:28 PM
P: 57
Hey guys, I know this is bringing up an old topic, but I wanted to inquire about something, as well as make sure I approached the final equation correctly.

To clear up the confusion from vela- yes, I am planning to go with the description saying that since they said to find the volume as indicated in the picture, my limits I set up was going to be from 0.

Therefore, I went ahead and said

∫∫(2x^2-y^2)-(4-y^2),y,0,√2-x^2),x,0,√2).
After evaluating this, I obtained -pi as my answer. the number makes sense, but the sign is wrong- negative volume is obviously impossible. When I reversed the two functions, I indeed get pi as the answer.

However, that doesn't make sense- wouldn't the z1 function have to be the function of the paraboloid, and the volume is a subtraction of the cylinder function z2 from z1?

Any clarification and a check to the final answer will be appreciated. Thanks guys!!
vela
#12
Jan16-12, 10:38 PM
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Quote Quote by iqjump123 View Post
However, that doesn't make sense- wouldn't the z1 function have to be the function of the paraboloid, and the volume is a subtraction of the cylinder function z2 from z1?
Why do you think that would be the case?


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