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Volume of a Prism

  1. May 22, 2008 #1
    Find the volume of the prism whose base is the area in the [tex]xy[/tex] plane bounded by the curves [tex]y = x^4[/tex] and [tex]y=x[/tex] and whose top is the surface [tex]f(x,y) = x+xy[/tex]

    this is what I have so far

    area of base = [tex]\int\ x^4dx - \int\ xdx[/tex]
    = [tex]$\frac{x^5}{5} - \frac{x^2}{2} + C[/tex]

    however I don't know what is meant by the top. I might guess it would represent the height of the prism and just multiply the expression by the one I have for the area, but that doesn't seem right, and this question is worth 10 marks
     
  2. jcsd
  3. May 22, 2008 #2

    Dick

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    It's not right. You find the volume by integrating the height over the region between the two curves.
     
  4. May 22, 2008 #3
    Hmm looks like my tex got a bit messed up on that first post, but it's not hard to see what I did there.

    What exactly does this mean? Is the f(x,y) function the height? Also, the region between the two curves is an area, how do I integrate over that?
     
  5. May 22, 2008 #4

    Dick

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    You integrate f(x,y)*dx*dy. The substance to the problem is figuring out what the x and y limits are. Draw a sketch of the region between the two curves.
     
  6. May 22, 2008 #5
    The enclosed region lies between (0,0) and (1,1) on the xy plane. So using 0 and 1 as limits for both x and y, this question simply becomes the double integral of x+xy? Doing this quickly I get an answer of 3/4, is this correct?
     
  7. May 22, 2008 #6

    Dick

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    Putting x and y limits both to 0 and 1 means you are integrating over a rectangle in the xy plane. That's not what you want. Let's take the x range from 0 to 1. Now pick a value of x. Look at your sketch and tell me what the corresponding y range is. It will depend on x.
     
  8. May 22, 2008 #7
    I thought that was too simple. Ok, so if for example I take [tex]x_0=0.5[/tex] then [tex]y_0[/tex] is in the range 1/16 to 1/2. What now? Since the shape is a prism, would it work if I took the volume produced by integrating over the rectangle, divided by the area of the rectangle and multiplied by the area of the bounded region? Though even if that were to work, it's probably not the method the examiners are looking for.
     
  9. May 22, 2008 #8

    Dick

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    No. Wouldn't work anyway. So you got if x=(1/2) then the y range is (1/2)^4 to (1/2). Good. So at a general value of x the y limits are x^4 to x, right? Put those as your dy limits.
     
  10. May 22, 2008 #9
    Omigosh I think it just clicked, thankyou. I first integrate with respect to y using x and x^4 as limits, treating each x as a constant. Putting in the limits gives me an expression entirely in x which I can then integrate with respect to x. Using 0 and 1 still for limits of x this gives me an answer of 29/120. have I got it this time?
     
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