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Finding volumes via double integrals

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume which lies below the plane [itex]z = 2x + 3y[/itex] and whose base in the [itex]x - y[/itex] plane is bounded by the [itex]x-[/itex] and [itex]y-[/itex]axes and the line [itex]x + y = 1[/itex].


    2. Relevant equations

    [itex]I = \int\int_{R} f(x, y) dydx = \int^{b}_{a}\int^{y=y_{2}(x)}_{y=y_{1}(x)} f(x, y) dydx[/itex]


    3. The attempt at a solution

    I know the integral is of the form:

    [itex]\int\int_{C} (2x + 3y) dydx[/itex]

    But I am very confused which what the limits are as they give so much information!

    I thought maybe something along the lines:

    [itex]x = 1 - y \Rightarrow 0 = 2(1 - y) + 3y = 2 - 2y + 3y = 2 + y \Rightarrow y = -2 \Rightarrow x = 3[/itex] (not sure if this helps in any way at all)

    And : [itex]2x + 3y = 0 \Rightarrow 3y = -2x \Rightarrow y = -\frac{2}{3}x[/itex] (but once again not sure if this even means anything)

    I haven't been formally taught this (at least finding volumes I haven't - I have had 2 tutorials on simply integrating over a region R) and expected to simply DO the assignment :(

    Thanks for any help!
     
  2. jcsd
  3. Jan 24, 2013 #2

    tms

    User Avatar

    Consider slicing the volume with vertical slices parallel to the [itex]y[/itex]-axis, which results in this integration:
    [tex]V = \int A(x) dx[/tex]
    where [itex]A(x)[/itex] is the area of each slice. To do the integration, first find an expression for [itex]A(x)[/itex] in term of [itex]x[/itex].
     
  4. Jan 24, 2013 #3
    I'm really sorry but I have no idea what you are asking for :(

    Would the limits be 0 <= x <= 1

    -x <= y <= x

    ?
     
  5. Jan 24, 2013 #4

    tms

    User Avatar

    Look up "volume by slicing".
     
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