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Volume of a Prism

  • Thread starter Gwilim
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  • #1
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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
 

Answers and Replies

  • #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.
 
  • #3
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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.

It's not right. You find the volume by integrating the height over the region between the two curves.
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?
 
  • #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.
 
  • #5
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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?
 
  • #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.
 
  • #7
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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.
 
  • #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.
 
  • #9
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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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