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Homework Help: Find volume polar coordinates

  1. Jul 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Compute the volume of the indicated solid

    Below z = sqrt(x^2+y^2), above z = 0, and inside x^2 + (y-1)^2 = 1


    2. Relevant equations

    3. The attempt at a solution

    My professor solved this in class but I didn't understand why deta is from -pi/2 to pi/2.

    It is obvious that the region D for integration is the trace of x^2+(y-1)^2 = 1
    In xy-plane it's a disk whose center is at (0,1)

    Since this is not a unit circle at (0.0), my professor said we have to express this disk in polar form, and I can't tell what the reason is. Maybe we want to find the range for the angle and radius?
    x^2 + (y-1)^2 = 1 in polar form

    x^2 + y^2 - 2y +1 - 1 =0 > r^2 -2rsin(x) = 0 where x is deta
    r^2 = 2rsinx
    r = 2sin(x) this is the polar form for x^2 + (y-1)^2 = 1

    Integration region D in polar form: {(r,x): 0 <= r <= 2sinx , -pi/2 <=x <= pi/2} and I know the integrand is r (polar form for sqrt(x^2+y^2) )

    Why is it -pi/2 and pi/2? How can I tell when the problem varies?

    Thank you.
     
  2. jcsd
  3. Jul 9, 2010 #2

    vela

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    It's relatively easy to express the integral for the volume in cylindrical coordinates. You could use Cartesian coordinates if you really wanted to, but why make things more difficult?
    You could let the angle go from 0 to pi, if that's your question, and you'd get the same answer. You just need an interval for the angle which will trace out the entire circle.
     
  4. Jul 9, 2010 #3
    Yes. But I would like to use polar before diving into clindrical.
    But how did you see it's 0 to pi, or -pi/2 to pi/2
    I can't figure it out.

    Thank you.
     
  5. Jul 9, 2010 #4

    vela

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    Well, to find a volume, you need a three-dimensional coordinate system, which means cylindrical, not polar, but cylindrical is just polar plus the z coordinate.

    To find the necessary range of θ, just plot the function r=2 sin θ, and see what values work.
     
  6. Jul 9, 2010 #5
    Hi thank you vela.
    I still don't know how to work out the values.

    Do you mean solve for sin(x)???
    Then what should i put for r? 1/2r = sin(x)? It shows pi/2 of course. But then what about -pi/2 , this will give us -1/2
     
  7. Jul 9, 2010 #6

    vela

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    Do you know how to plot a polar function r=r(θ)?
     
  8. Jul 9, 2010 #7
    I can draw r = 2sin(x) :)
    just make a table with x, and r = 2sin(x) and the polar coordinates label just like rectangular (0,1,2,3,4,5...)

    x = 0, r = 2sin(x) is 0
    x = pi/2, r = 2
    x = pi, r = 0
    x = 3pi/2, r = -2
    x = 2pi, r = 0

    if we plot from 0 to 2pi, we get a graph that looks like "8"
    from 0 to pi, we get exactly what we have in rectangular, x^2 + (y-1)^2 = 1, whose center is at (0,1) with radius = 1

    but how is it -pi/2 and pi/2

    i don't think 0 to pi will give me the same answer as from -pi/2 to pi/2

    answer from wolframalpha
    integrate r^2 dr dx, x = 0 to pi, r = 0 to (2*sin(x)) -> 32/9
    integrate r^2 dr dx, x = -pi/2 to pi/2, r = 0 to (2*sin(x)) -> 0


    This is the solid...
    vgr290.jpg
     
    Last edited: Jul 9, 2010
  9. Jul 9, 2010 #8

    vela

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    Try plotting the function letting θ run from -pi/2 to pi/2. You'll get the same circle.
    Why are you integrating r2 dr dθ? That's not the volume element in cylindrical coordinates.
     
    Last edited: Jul 9, 2010
  10. Jul 9, 2010 #9
    thank you for the response.
    vela i am not doing it with triple integration.
    i have to finish this question using double integrals...

    even if i am doing it with triple integrals, i still need to figure out region D (the most two outer integrals)...

    Compute the volume of the indicated solid
    Below z = sqrt(x^2+y^2), above z = 0, and inside x^2 + (y-1)^2 = 1

    so i thought the integrand would be sqrt(x^2+y^2) which in polar is r

    integral ( alpha to beta) integral (a to b) of integrand * r dr d (theta) over region D
    where the region D in polar is {(r,theta): r b/w a and b, theta b/w alpha and beta }

    so instead of writing r r dr d theta, i just wrote r^2 dr dtheta for double integrals

    triple integrals for volume is triple integrals 1 dv in general....
     
  11. Jul 9, 2010 #10

    vela

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    Yes, you're right. I had done it as a triple integral in Mathematica, and I get the same results for both ranges of angle. Let me look into what's going on a bit more.
     
  12. Jul 9, 2010 #11

    vela

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    There's a complication from using the interval -pi/2 to pi/2 for θ because from -pi/2 to 0, r=2 sin θ is negative. That means the height of the volume z=sqrt(x2+y2) isn't simply r, but |r|. The integrand should therefore be r |r|. If instead you integrate from 0 to pi, r will be always be positive, so you can simply integrate r2 to get the answer.

    Considering this, it seems strange your professor would opt to use -pi/2 to pi/2 instead of 0 to pi unless it was to illustrate this point.
     
  13. Jul 10, 2010 #12
    I think you should draw a prettier picture. Not hard to do in Mathematica which personally I feel should be an integral part of any Calculus course:

    cp1 = ContourPlot3D[{z == Sqrt[x^2 + y^2], x^2 + (y - 1)^2 == 1,
    z == 0}, {x, -3, 3}, {y, -3, 3}, {z, 0, 3},
    ContourStyle -> {LightPurple, Opacity[0.4]}, Mesh -> None,
    AxesLabel -> {"x", "y", "z"}]

    Note in the plot below, we're looking at it with the x-axis at the left side. Now, you derived the expression for r as a function of t: [itex]r=2\sin(t)[/itex] and from the picture, t clearly goes from zero to pi (since the circle is above the x-axis) or just 2 times from zero to pi/2 since it's symmetrical on both sides. So looks to me it's just:

    [tex]V=2\int_0^{\pi/2}\int_{0}^{2\sin(t)} (r) rdrdt[/tex]

    I think the -pi/2 thing was not correct.
     

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