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Double integral question.

  1. Dec 18, 2012 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I am asked to find the volume under the curve whose equation is z=16-(x^4+y^4), and within (x^2+y^2)<=1, using a double integral.


    2. Relevant equations



    3. The attempt at a solution
    Should I use cylindrical coordinates?
    I feel slightly lost. I have tried drawing z=16-(x^4+y^4), unsuccessfully.
    I understand that (x^2+y^2)<=1 is a circle of r<=1 and sets the limit to the volume, i.e. the base of the shape.
    I'd truly appreciate some guidance.
     
  2. jcsd
  3. Dec 18, 2012 #2

    sharks

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    Here's the plot for ##z=16-(x^4+y^4)##:
    http://img8.uploadhouse.com/fileuploads/17053/17053388c9a8698ea1f3d43b18b267345d128d42.gif
    Cylindrical coordinates should work fine. First, convert the equations of the curve and circle into polar coordinates.
     
  4. Dec 18, 2012 #3

    Dick

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    These are surfaces, not curves. It's sort of a boxy 'paraboloid like thing' opening down. You really don't need to know much about the detailed shape. The important thing about it is that it lies completely above the disk x^2+y^2<=1 in the xy plane. Having said that I now realize I was assuming the xy plane was the lower bound. x^2+y^2<=1 is actually the inside of an infinite cylinder. If there is not some other z cutoff the volume will be infinite. Is that exactly how the problem is stated?
     
  5. Dec 18, 2012 #4

    sharks

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    That's exactly what i thought, but then there would be no purpose to calculating the volume, as the answer would be directly obtained theoretically. I believe it's a wrongly worded problem and the plane ##z=0## should be the lower limit. The OP should have provided the complete question accurately.
     
  6. Dec 18, 2012 #5
    The volume bounded under the tent-like shape given by the equation z=16-(x^4+y^4), and within the area (x^2+y^2)<=1.
    Is it clearer now?
     
  7. Dec 18, 2012 #6

    Dick

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    z=16-(x^4+y^4) defines the top, the cylinder x^2+y^2=1 defines the sides and the bottom is defined by z=0 i.e the xy plane. Is that what you mean? The 'bottom' needs to somehow define the lowest value of z.
     
  8. Dec 18, 2012 #7
    So z changes from 0 to 16-(x^4+y^4) while y changes from 0 to sqrt(1-x^2) (the volume then to be multiplied by two)?
     
  9. Dec 18, 2012 #8

    SammyS

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    @peripatein,

    All the respondents are asking you, "Is the region bounded by z=0 below?" You're the one with access to the problem that you're trying to solve.
     
  10. Dec 18, 2012 #9
    But I have already stated the problem exactly as it is presented to me, verbatim! I am in possession of no further information.
     
  11. Dec 18, 2012 #10

    sharks

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    In that case, you should inform your teacher about the missing lower bound for z. If it's a problem you got from a book, it is missing some required information.

    Anyway, when you get the equation of the lower plane (we had assumed it was ##z=0##), just use that value as the lower limit of ##z## in your double integral.
     
  12. Dec 18, 2012 #11
    May I, in that case, say that z changes from 0 to 16-(x^4+y^4) while y changes from 0 to sqrt(1-x^2) (in which case the volume would then have to be multiplied by 2)?
     
  13. Dec 18, 2012 #12

    sharks

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    If you are evaluating the volume using Cartesian coordinates, you need to specify the limits for ##x## as well.

    Using Cartesian coordinates, if you say that x varies from 0 to ##\sqrt (1-y^2)##, the volume obtained after evaluating the triple integral would need to be multiplied by 4, since you would then have calculated the volume of the solid in the first octant only. But if you state that x varies from ##-\sqrt (1-y^2)## to ##\sqrt (1-y^2)##, then in that case, you would have to multiply by 2, in order to get the final volume.

    But it would be easier to use cylindrical coordinates. Refer to post #2.
     
    Last edited: Dec 18, 2012
  14. Dec 18, 2012 #13
    As I am requested to use a double integral, using cylindrical coordinates should r change from 0 to 1 while z changes from 0 to 16-(x^4+y^4), with Jacobian of transformation r(dr)(dtheta)(dz)? Or do I need to let theta change from 0 to 2pi? I am not sure.
     
  15. Dec 18, 2012 #14

    sharks

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    You need to provide the limits for the following 3 parameters: r, θ and z.
    If θ varies from 0 to 2∏, you'll get a complete revolution about the z-axis, forming the entire volume of the solid.
    You also need to re-write the equation of the curve in terms of polar coordinates before you proceed with evaluating the triple integral.
     
  16. Dec 18, 2012 #15

    SammyS

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    You are referring to 3 variables here, r, θ, and z.

    To find the volume with a double integral rather than a triple integral, you need to have an integrand that's different than 1.

    [STRIKE]The obvious integrand, in my opinion, is 16-r4, and integrate over r and θ, using the appropriate Jacobian.

    Alternatively, you could use an integrand of 2π r and integrate over r and z.
    [/STRIKE]


    Added in Edit:
    I've crossed out the baloney !
     
    Last edited: Dec 19, 2012
  17. Dec 19, 2012 #16
    Is z not equal to 16-r^4(cos(2theta))^2, where r is <= 1?
     
  18. Dec 19, 2012 #17

    sharks

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    ##x^4+y^4=r^4## is easily obtained from the regular expression ##x^2+y^2=r^2##, if you substitute ##x=x^2## and ##y=y^2## into the previous formula.

    Since ##z=16-(x^4+y^4)## is now ##z=16-r^4##, at ##r=1##, what is the value of ##z##? To find the maximum value of ##z##, set ##r=0##, which is the origin.
     
  19. Dec 19, 2012 #18
    It is still not very clear to me. First of all, why is that substitution (namely, x=x^2, y=y^2) even valid? Second, if z changes from 15 to 16, do I substitute these numerical values or do I use parameters still? What should be substituted for the (dz) element of the Jacobian? (-4r^3)dr?
     
  20. Dec 19, 2012 #19

    CAF123

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    I am a little confused here too. If $$(x^2 + y^2) = r^2,\, \text{then}\, (x^2 + y^2)^2 = (r^2)^2.$$ This implies $$r^4 = (x^2 + y^2)^2 \neq x^4 + y^4?$$
     
  21. Dec 19, 2012 #20

    CAF123

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    So you are effectively saying that ##cos^4x + sin^4x = 1?## While this holds for ##x=0##, it does not for ##x = \pi/4##.

    http://www.wolframalpha.com/input/?i=cos^4x+++sin^4x

    EDIT: Just so anyone else who reads this does not get confused, this was in response to a post that has now been deleted.
     
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