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Interpret the integrand

  1. Aug 28, 2014 #1

    Zondrina

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

    Interpret the integral as the volume of a certain solid and describe the solid geometrically. Calculate the volume.

    $$\int_0^a dy \int_0^{\sqrt{a^2 - y^2}} \frac{2x + 4y}{3} dx$$

    2. Relevant equations



    3. The attempt at a solution

    Clearly we have an upper surface ##z = \frac{1}{3} (2x + 4y)##, which is a plane through the origin since ##(x,y) = (0,0) \Rightarrow z = 0##.

    The region appears to be described by:

    $$R := \{(x,y) \in \mathbb{R^2} \space | \space x=0, y=0, x^2 + y^2 = a^2 \}$$

    So we have a cylinder of radius ##a##, which we have sliced by the planes ##x = 0## and ##y = 0##. This leaves a quarter of a cake piece in the first quadrant.

    Then the plane ##z = \frac{1}{3} (2x + 4y)## cuts a chunk out of the bottom half of the cake, creating a quarter cake piece with a slope on its top face aimed towards the origin.

    I'm not sure what you would call this solid. Calculating the volume of the integral is easy afterwards.

    EDIT: It seems like a sloped quarter disk?
     
    Last edited: Aug 28, 2014
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  3. Aug 28, 2014 #2

    haruspex

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    OK, but what is the lower surface?
    You don't want "=" in there.
     
  4. Aug 28, 2014 #3

    Zondrina

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    The lower surface is a cylinder. I should have mentioned that. Also, that '=' should read '≤'.
     
  5. Aug 28, 2014 #4

    haruspex

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    You're trying to express the integral as the volume of a solid. That means it should correspond to a triple integral ∫∫∫dz.dy.dx over some region. But you are given ∫∫(2x+4y)/3 dy.dx. How do you make that look like a triple integral of "1"?
     
  6. Aug 28, 2014 #5

    Zondrina

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    In that context we should talk in generality for a moment. Suppose ##S## is the 3-D region in question then the volume of ##S## is given by:

    $$V_S = \int \int \int_S dV$$

    We know that ##S## is the region below ##z = \frac{1}{3} (2x + 4y)## and above the region ##R## in the x-y plane. So we can re-write the integral:

    $$V_S = \int \int \int_S dV = \int \int_R \left[\int_0^z dz \right]dA = \int \int_R z dA = \int \int_R \frac{1}{3} (2x + 4y) dA$$

    EDIT: Just to be a little more rigorous, I should mention:

    $$S := \{ (x,y,z) \space | \space (x,y) \in R, \space x=0, \space y = 0, \space x^2 + y^2 \leq a^2, \space 0 \leq z \leq \frac{1}{3} (2x + 4y) \}$$
     
    Last edited: Aug 28, 2014
  7. Aug 28, 2014 #6

    haruspex

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    Right, this is the lower surface: 0 ≤ z.
    Note, though, that this is arbitrary. You could have validly chosen 1 ≤ z ≤ (2x + 4y)/3 + 1, etc.
     
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