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Volume of a Solid- 3D Calculus

  1. Apr 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the given solid.
    Bounded by the planes
    z = x, y = x, x + y = 5 and z = 0

    2. Relevant equations
    V = [PLAIN]http://www.webassign.net/wastatic/wacache6cdd60ea0045eb7a6ec44c54d29ed402/watex/img/integral.gif[PLAIN]http://www.webassign.net/wastatic/wacache6cdd60ea0045eb7a6ec44c54d29ed402/watex/img/integral.giff(x,y) [Broken] dA
    For a type I integral: dA=dy*dx
    and the domain is: a<x<b, g1(x)<y<g2(x)

    For a type II integral: dA=dx*dy
    and the domain is: c<y<d, h1(y)<x<h2(y)

    3. The attempt at a solution

    I chose to evaluate this integral by making region D as a type I integral, therefore my domain is as follows:
    0<x<5/2 & x<y<5-x
    My f(x,y)=x
    dA=dy*dx

    Is it possible to do this as a type II integral? That way my limits would look a lot nicer, but I don't know what my h1(y) and h2(y) functions would be.
     

    Attached Files:

    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 1, 2016 #2
    Hi fridakahlo:

    I think that when you calculate a volume by integration you need an integral of the form
    V = ∫∫∫ f(x,y,z) dV = = ∫∫∫ f(x,y,z) dx dy dz

    Hope this helps.

    Regards,
    Buzz
     
  4. May 1, 2016 #3

    haruspex

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    No, a volume integral would just be ∫∫∫dxdydz, no need for a function. But in this case fridakahlo is effectively assuming the z integral has been done, leaving either ∫∫fdxdy or ∫∫dydx.

    @fridakahlo , yes you can do the x integral before y, but you would need to split the outermost integral into two ranges, y<2.5 and y>2.5.
     
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