Calc II Problem

  • Thread starter Spectre32
  • Start date
  • #1
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Ok i'm working on this Integral: http://home.comcast.net/~bruins83/IMG_1527.jpg [Broken]

I figured the Integral to be equal to 11/2. Now they want me to graph it, and I'm slidght confused on how to go about this, The direction read, Sketch the solid whose volume is given by (integral) Find the volume.

Any help would be appericated.

Thanks
 
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Answers and Replies

  • #2
1,444
2
i can't see what you've posted there properly can you just type it out
 
  • #3
dav2008
Gold Member
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[tex]\int_{0}^{1} \int_{0}^{1}2-x-y~dydx[/tex]
 
  • #4
1,444
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dav2008 said:
[tex]\int_{0}^{1} \int_{0}^{1}2-x-y~dydx[/tex]
the solution to that is just one

the new function after evaluating the indefinite integral is [tex]\frac{-1}{2} xy (x+y-4) [/tex]
 
  • #5
136
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Yeha I went back over it, and got one, sorry about that. Anyways, I still need help graphing it. Actually what DO i even look at to graph?
 
  • #6
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Spectre32 said:
Yeha I went back over it, and got one, sorry about that. Anyways, I still need help graphing it. Actually what DO i even look at to graph?

if you want to graph (sketch) it then you have to hold Y constant and look at all the ways x can move around the axes. After that simply enclose the surface

it loos like a gentle slope down a hill and then a little rise on the positive side
 
Last edited:
  • #7
dav2008
Gold Member
609
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What exactly do they want you to graph? Can you quote the book exactly?

If they just want you to draw in the region over which the function is being integrated then you just take the limits and graph them on the x/y plane.

Your limits are y=0, y=1, x=0, x=1 so it would just be a square.

Edit: Never mind, I reread your post and it says to sketch the solid.
 

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