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Finding volume bounded by paraboloid and cylinder |
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| Jul1-11, 10:13 AM | #1 |
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Finding volume bounded by paraboloid and cylinder
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! |
| Jul1-11, 12:54 PM | #2 |
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Are your equations correct because z=4-y2 isn't the equation of a cylinder?
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| Jul1-11, 03:53 PM | #3 |
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| Jul1-11, 04:56 PM | #4 |
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Finding volume bounded by paraboloid and cylinder
D'oh!
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| Jul4-11, 11:09 PM | #5 |
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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! |
| Jul5-11, 12:06 AM | #6 |
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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] |
| Jul6-11, 11:39 PM | #7 |
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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! |
| Jul7-11, 12:48 AM | #8 |
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Why are the lower limits 0 for both x and y?
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| Jul7-11, 07:21 AM | #9 |
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| Jul7-11, 11:34 AM | #10 |
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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.
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| Jan15-12, 08:28 PM | #11 |
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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!! |
| Jan16-12, 10:38 PM | #12 |
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| cylinder, integral, paraboloid, volume |
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