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Bounded regions and triple integrals

  1. Dec 6, 2014 #1
    1. The problem statement, all variables and given/known data

    a) sketch the region in the first octant bounded by the elliptic cylinder 2x^2+y^2=1 and the plane y+z=1.
    b) find the volume of this solid by triple integration.
    2. Relevant equations


    3. The attempt at a solution
    I have already sketched the elliptic cylinder and the plane. my problem arises with the triple integration.
    I do not understand if i have to find the volume of the entire solid or the volume of the solid in the first quadrant.
    also i am having trouble setting up the integrals because of the ellipse. i tried both cylindrical coordinates and spherical but i get stuck because of the 2.

    for spherical coordinates i get stuck at the following: rho^2sin^2(phi)(2cos^2(theta)+sin^2(theta))=1
    for cylindrical i get stuck here: 2r^2cos^2(theta)+r^2sin^2(theta)=1
     
  2. jcsd
  3. Dec 6, 2014 #2
    i have also figuired the following integral out but im not sure if it correct. i did this one in rectangular coordinates.

    4(∫dx∫dy∫dz)

    where the limits of integration for x are 0 to 1/√2
    for y are 0 to √(1-2x^2)
    for z are 0 to 1-y

    can anybody comment on this?
     
  4. Dec 6, 2014 #3

    Zondrina

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    You want to find the volume of the solid under ##z = 1 - y## and above the elliptic cylinder ##2x^2 + y^2 = 1## in the first octant, i.e ##x, y, z \geq 0##.

    $$\iiint_V \space dV = \iint_R \int_0^{1-y} \space dzdA = \iint_R 1 - y \space dA$$

    Now project the elliptic cylinder onto the x-y plane, this will allow you to obtain limits for ##x## and ##y##. To obtain these limits, I would suggest creating an invertible transformation between the ##(x,y)## and ##(u,v)## space and then finding the corresponding Jacobian.
     
  5. Dec 6, 2014 #4
    i dont know how to do the transformations; owever, i have come up with the integral, the problem is i dont know how to compute it. i came up with the integral in rectangular form and it goes as follows
    ∫dx∫dy∫dz

    where the limits of integration for x are 0 to 1/√2
    for y are 0 to √(1-2x^2)
    for z are 0 to 1-y

    also i tried to convert cylindrical and spherical coordinates but was then stuck becasue of the two, any help on how to compute this integral would be greatly appreciated.
     
  6. Dec 6, 2014 #5

    Zondrina

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    Your Cartesian integral looks fine, and can be handled with simple substitutions.

    The reason I suggested the ##(u,v)## co-ordinate transform though is it will allow you to map the elliptic cylinder in ##(x,y)## space to a circle of radius 1 in the ##(u,v)## space. Now that you have a ##(u,v)## integral, you can apply polar co-ordinates to the circle and basically read limits off without thinking.

    In case you wanna try it, let ##x = \frac{1}{\sqrt{2}} u## and ##y = v##, the elliptic cylinder then maps to ##u^2 + v^2 = 1##.
     
  7. Dec 6, 2014 #6
    how can i solve the integral with simple subsittution? if i break the integral up into 2 diferent integrals and use the sum rule, in one integral i have a square root and u substituion does not hold for this case as far as i can tell, so how would i go about solving this problem?
     
  8. Dec 6, 2014 #7

    Zondrina

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    You have the integral:

    $$\iint_R 1 - y \space dA = \int_0^{\frac{1}{\sqrt{2}}} \int_0^{\sqrt{1 - 2x^2}} 1 - y \space dydx = \int_0^{\frac{1}{\sqrt{2}}} \sqrt{1 - 2x^2} - \frac{1 - 2x^2}{2} \space dx$$

    The integral of ##\sqrt{1 - 2x^2}## should remind you of a first year integral. It is of the form ##\sqrt{a^2 - y^2}## and should ring bells about trig substitutions.
     
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