Finding volume bounded by paraboloid and cylinderby iqjump123 Tags: cylinder, integral, paraboloid, volume 

#1
Jul111, 10:13 AM

P: 57

1. The problem statement, all variables and given/known data
Find the volume bounded by the paraboloid z= 2x^{2}+y^{2} and the cylinder z=4y^{2}. Diagram is included that shows the shapes overlaying one another, with coordinates at intersections. (Will be given if necessary) 2. Relevant equations double integral? function1function2? 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! 



#2
Jul111, 12:54 PM

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Are your equations correct because z=4y^{2} isn't the equation of a cylinder?




#3
Jul111, 03:53 PM

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#4
Jul111, 04:56 PM

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Finding volume bounded by paraboloid and cylinder
D'oh!




#5
Jul411, 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! 



#6
Jul511, 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] 



#7
Jul611, 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(2x^2), and dx will stretch from 0 to sqrt(2). Is this correct? Thanks! 



#8
Jul711, 12:48 AM

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Why are the lower limits 0 for both x and y?




#9
Jul711, 07:21 AM

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#10
Jul711, 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.




#11
Jan1512, 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^2y^2)(4y^2),y,0,√2x^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!! 



#12
Jan1612, 10:38 PM

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