anonymous
  • anonymous
What is the volume of y=sqrt(x), y=-x, y=16, with cross sections of isosceles right triangles with on leg on the base??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I'm confused as to what the "base" is? Can you describe the problem a little more?
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anonymous
  • anonymous
Well i am given the y=sqrt(x), y=-x, y=16, and i picked the base as right isosceles triangles with one leg on base, and i need to make a 3-D model of the solid.
anonymous
  • anonymous
|dw:1332652896253:dw|

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anonymous
  • anonymous
Okay, I'm seeing it now. Is the leg that is in the "base" running vertically or horizontally?
anonymous
  • anonymous
perpendicular to the x-axis
anonymous
  • anonymous
Aight. In that case, we need to split the area into two areas. We have to compute separately the volume for the part in the second quadrant and another for the part in the first quadrant.
anonymous
  • anonymous
Starting with the part in the second quadrant, the length of the base is 16 - (-x) = 16 + x. As such, the area of the cross section is A(x) = 0.5(16 + x)^2
anonymous
  • anonymous
anonymous
  • anonymous
Then, we have a generic volume formula with \[V=\int\limits_{a}^{b}A(x)dx=\int\limits_{-16}^{0}1/2(16+x)^2dx\]. Do you see how I got that? We are integrating from x = -16 to x=0
anonymous
  • anonymous
yes i got 642.667 for the volume
anonymous
  • anonymous
You sure? I seem to be getting 682.667. Anyway, then we do the same thing for the first quadrant portion. This time, the base of the yellow triangle above is 16 - sqrt(x), giving A(x) = 0.5(16 - sqrt(x))^2
anonymous
  • anonymous
\[V=\int\limits_{0}^{256}0.5(16-\sqrt{x})^2dx\] will give the volume of the right side. Add this to 682.667 to get the total volume. Do you understand where the limits of integration came from?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Okay. Well, summing the volumes, I get 6144 as the volume of the solid.
anonymous
  • anonymous
yes, how would i individually find the area and volumes of cuts? I added the 256+16 to get 272 and divided by 16 to get 17 so that there would be 17 cuts, is this correct?
anonymous
  • anonymous
Ok, tell me if this is correct for finding area and volume of one slice. 272(2/17)=32 so A=.5(16+32)^2 = 1152, then \[v=\int\limits_{-16}^{0}.5(16+1152)^2\]
anonymous
  • anonymous
Well, you should identify which slice you're finding the volume for. If you use x = 32, then you ought to use the A = 0.5(16 - sqrt(x))^2 formula becaue you are computing a slice in the first quadrant. Second, you plug A into the volume formula where there's an A, NOT where there's an x.
anonymous
  • anonymous
But yes, this procedure will give you a rectangular prism that approximates the volume of the slice of the solid. Just make sure that you know which slice you're doing, very important.
anonymous
  • anonymous
How would I find slices in the second quadrant? I am confused and stuck
anonymous
  • anonymous
stuck :'(

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