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Estimating the volume of a solid

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data
    estimate the volume of the solid
    z=-2(x^2+y^2)+8
    between the two plates z=4 and z=0


    2. Relevant equations
    In question like this, should I use triple integrals or double integrals in polar coordinates? I'm stuck in between which to use, because the question asks to estimate the volume which suggest a triple integrals. Yet the function gives me a strong feeling that I should use a double integrals in polar coordinates instead. Any suggestions?

    Even though I'm not sure have I done it right, I tried solving it using double integrals in polar coordinates. Please check have I done it right, and if I should do it in triple integrals do please give me a guideline of how to do it (as it has not been taught to us yet, but the assignment is due before our next class)

    p={(r,θ)= 0≤ r ≤4, 0≤ θ ≤ [itex]\pi[/itex]

    ∫∫[itex]_{p}[/itex] -2(x2+y2)+8=∫[itex]^{\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]

    3. The attempt at a solution

    ∫∫[itex]_{p}[/itex] -2(x2+y2)+8=∫[itex]^{\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]-2(r2)r Δr Δθ

    ∫[itex]^{\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex] -2r3 Δr Δθ

    ∫[itex]^{\pi}_{0}[/itex] [-2r4/4][itex]^{4}_{0}[/itex] Δθ

    ∫[itex]^{\pi}_{0}[/itex][-2(4)4/4]-[-2(0)4/4] Δθ

    ∫[itex]^{\pi}_{0}[/itex]-128 Δθ

    [-128θ][itex]^{\pi}_{0}[/itex]

    =-128[itex]\pi[/itex]

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 25, 2012 #2

    Bacle2

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    I haven't gone thru the whole thing yet, but the minus sign in the answer is

    a sign of a problem somewhere--notice you dropped the 8 when you did your change to polars. Still, the figure you get is the frustum* of a cone:

    http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html to help you

    double check.

    *Also the name of my favorite drink!

    EDIT: I misread the problem, sorry. Ultimately,you can use dimensional analysis to

    decide if you need a double- or triple- integral. A single integral will do if you have

    the general area in the integrand.
     
    Last edited: Aug 25, 2012
  4. Aug 25, 2012 #3
    ok, but even if I have the +8 in there, shouldn't still gives me a negative number?

    secondly should I keep using polar or should I really use triple integration?

    thanks
     
  5. Aug 25, 2012 #4

    HallsofIvy

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    May I ask why you use the word "estimate" rather than "calculate"? Because of the circular symmetry, I would recommend using cylindrical coordinates. That is, three dimensions, but using polar coordinates in place of x and y.

    Bacle2, this is NOT the frustrum of a cone. That would be true if it were [itex]z^2= 8- 2(x^2+ y^2)[/itex]. This figure is a 'frustrum' of a paraboloid.
     
  6. Aug 25, 2012 #5

    LCKurtz

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    Best is a triple integral in cylindrical coordinates. See below

    That isn't set up correctly. If you are going to use a double integral in cylindrical coordinates the integrand would be ##z_{upper}-z_{lower}##. You have used the equation of the paraboloid, but the paraboloid is cut off at the top by the plane ##z=4##. The upper surface is partly the paraboloid and partly the plane, which is a good reason not to do it that way. Also, ##r## doesn't go from 0 to 4. The largest ##r## is is 2 at the base of the solid. The values ##r## takes on depend on the value of ##z##.

    My suggestion would be to try to set it up as a triple integral in the order ##drdzd\theta##. You will have to get ##r## in terms of ##z## for the upper limit, but the rest will be easy.
     
  7. Aug 25, 2012 #6

    ok, I shall give that a try. but how should I do the ##dz## part?
    Do I first rearrange the equation to ##z+2(x^2+y^2)-8## and then integrate from there?

    Phil
     
    Last edited: Aug 25, 2012
  8. Aug 25, 2012 #7
    well the assignment question used estimate.
    so can you tell me what's the difference between estimate and calculate in the context of this question?

    thanks,
    Phil
     
  9. Aug 26, 2012 #8

    LCKurtz

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    Instead of asking me how to do it, show what you think it should be. The dr part would be first. Have you done that? Have you drawn a picture of the figure?
     
  10. Aug 26, 2012 #9
    yes you are right, I should have tried it first.
    I'm not too sure what do you meant by values of ##r## takes on depend on the value of ##z##. But this is what I think you meant, and my attempt to solve it.
    ##z= -2(x^2+y^2)+8##
    ##z= -2r^2+8##

    ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]∫[itex]^{2}_{0}[/itex](##r cos\theta,rsin\theta,z)r## ##drdzd\theta##
    ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]∫[itex]^{2}_{0}[/itex](##r cos\theta,rsin\theta,-2r^2+8)r## ##drdzd\theta##

    ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]∫[itex]^{2}_{0}[/itex](##r^2 cos\theta,r^2sin\theta,-2r^3+8r)## ##drdzd\theta##

    ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]##[(r^3/3)*cos\theta,(r^3/3)*sin\theta,(-2r^4/4)+4r^2]##[itex]^{r=2}_{r=0}[/itex]##dzd\theta##

    ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]##[6/3cos\theta,6/3\theta,(-32/4)+16] dzd\theta##

    ∫[itex]^{2\pi}_{0}[/itex]∫[itex]^{4}_{0}[/itex]##[6/3cos\theta,6/3\theta,8] dzd\theta##

    ∫[itex]^{2\pi}_{0}[/itex]##[6/3 z cos\theta, 6/3 z sin\theta, 8z]##[itex]^{z=4}_{z=0}[/itex]##d\theta##

    ∫[itex]^{2\pi}_{0}[/itex]##[24/3cos\theta,24/3sin\theta,32]d\theta##

    ∫[itex]^{2\pi}_{0}[/itex]##[8cos\theta,8sin\theta,32]d\theta##

    ##=[8sin\theta,-8cos\theta,32\theta]##[itex]^{2\pi}_{0}[/itex]
    ##=[8sin(2\pi),-8cos(2\pi),32(2\pi)]##
    ##=[8sin(2\pi),-8cos(2\pi),64\pi]##
    ##=[0,-8,201]##

    This is what I think, but I believe there is something wrong, because the results is giving a set of points rather then a numerical value for volume.
     
    Last edited: Aug 26, 2012
  11. Aug 26, 2012 #10

    Bacle2

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    Why are you integrating a generic function of r, θ and z ? Again, think of

    dimensional analysis.


    My apologies again for unknowingly misleading you by misreading z for z2,

    believing the figure was a cone, instead of a paraboloid.
     
  12. Aug 26, 2012 #11

    LCKurtz

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    No, no, no! In a triple integral, volume is always given by$$
    \iiint_V 1\, dV$$The integrand is always 1 and the only problem is the limits.

    And I wonder if you even read my first post. Go back and look at it again regarding the limits, ##r## especially. What is ##r## in terms of ##z##? The limits you have put on would describe a right circular cylinder.
     
  13. Aug 27, 2012 #12
    Sorry for hijacking the thread, but I'm curious as to know how do you tell the shape of a solid/curve?

    I know common ones are:

    Sphere: x2 + y2 + z2 = 1

    Ellipsoid/nutshell: (x/a)2 + (y/b)2 + (z/c)2 = 1

    I am also very poor at doing problems involving coordinate geometry: parabolas, hyperbolas and ellipses especially..

    Any tips on improving my skills?
     
  14. Aug 27, 2012 #13
    what do you meant by think of dimensional analysis?

    I've read your post prior to my attempt, I think I did mention before my attempt that I do not understands what do you meant "what is ##r## in terms of ##z##" part. Can you please elaborate on that part more, and the part regarding on the limits.

    As I said this topic has not been taught yet, but the assignment is due prior to the next class. So my understanding to the topic as a whole is still vague.

    Phil
     
    Last edited: Aug 27, 2012
  15. Aug 27, 2012 #14

    SammyS

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    It seems to me that if you are merely expected to estimate the volume, then this has all been overkill.

    For any fixed value of z, from 0 to 4, the resulting equation is that of a circle.
    Solving for x2 + y2 gives,

    x2 + y2 = (8-z)/2​

    If z = 0, what is the area of the circle?

    If z = 1, what is the area of the circle?
    ...

    If z = 4, what is the area of the circle?
     
  16. Aug 27, 2012 #15

    LCKurtz

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    I have some old hand-written lecture notes for review of conics (Lecture 7) and how to use that to help you draw 3-D surfaces (Lecture 8). My handwriting isn't that great, but you might find them useful. Here's a couple of links:

    [/PLAIN] [Broken]
    http://math.asu.edu/~kurtz/LectureNotes/Lecture7.pdf[/URL] [Broken]

    [/PLAIN] [Broken]
    http://math.asu.edu/~kurtz/LectureNotes/Lecture8.pdf[/URL] [Broken]
     
    Last edited by a moderator: May 6, 2017
  17. Aug 27, 2012 #16

    LCKurtz

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    You figured out that ##z = -2r^2+8##. That tells you how ##z## depends on ##r##. If you solve it for ##r## you will have how ##r## depends on ##z##. ##r## goes from ##0## to the ##r## on the surface which is given by that equation. That gives you the upper limit for ##r##.

    But, given that this topic hasn't been taught yet, SammyS may have it right that you were just supposed to estimate the volume in the first place.
     
  18. Aug 28, 2012 #17

    Bacle2

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    I mean that , e.g., a volume being a measure of 3-dimensional content should be the

    product of either three linear measures, like (Length)(Width)(Depth) or the product

    of a 2-dimensional measure by a 1-d measure, like (Area of base)(Height) .
     
  19. Aug 28, 2012 #18
    Ok, I'm kinda catching on. So z has two conditions, which is z =4 and z=0.
    In order to work out the limits of r, I need to rearrange my equation making r the subject.
    hence one would be z=0, therefore ##2r^2=8## which makes ##r=2## as its lower limits.
    the other one would be z=4, therefore ##2r^2=4##, which makes ##r=√2=1##

    and then from here I do my integration, using ∫∫∫##-2r^2+8 drdzd\theta##. Though I still have some doubt that I should integrate using ##-2r^2+8##, but this is the only known equation to do my math
     
  20. Aug 28, 2012 #19

    LCKurtz

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    No. Almost everything you have written above is wrong or inappropriate for the problem. I think this is unproductive trying to explain it to you when you haven't had the material yet. And you haven't responded to any of the comments asking whether you are really meant to just estimate the volume instead of calculate it. Is English your natural language? Do you understand the difference between estimate and calculate?
     
  21. Aug 28, 2012 #20
    Why yes I meant estimate instead of calculate. And I believe I know the difference between estimate and calculate in terms of English. Though I'm not too sure I correctly understand the difference in terms of calculus.

    I'll come back to this question once I had the material taught.
     
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