Finding volume bounded by paraboloid and cylinder

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SUMMARY

The discussion focuses on calculating the volume bounded by the paraboloid defined by the equation z = 2x² + y² and the parabolic cylinder given by z = 4 - y². Participants clarify that the latter is indeed a parabolic cylinder, extending infinitely in the x-direction. The correct setup for the double integral is established as V = ∫∫(2x² + y² - (4 - y²)) dy dx, with bounds for y from 0 to √(2 - x²) and for x from 0 to √2. The final volume calculation yields a positive result of π after correcting the order of the functions in the integral.

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iqjump123
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Homework Statement


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)


Homework Equations


double integral? function1-function2?


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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Are your equations correct because z=4-y2 isn't the equation of a cylinder?
 
vela said:
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.
 
D'oh!
 
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!
 
As you mentioned in your original post, you want to calculate something like
V = \iint\limits_A [z_1(x,y)-z_2(x,y)]\,dy\,dx
 
Last edited by a moderator:
Why are the lower limits 0 for both x and y?
 
vela said:
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?
 
  • #10
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
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!
 
  • #12
iqjump123 said:
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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