Volume of solid

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volume of solid:(

hi..I tried to solve it but ı couldnt .book says the answer is pi /6...please help me.

question is; y=x and y=x^1/2 about y =1
 

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  • #2
arildno
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Pinpoint your problem.
 
  • #3
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ok I showed my work..


I used this formula
integral Pı(outer r)^2-Pı(inner)^2



inside integral from y=0 to y=1 pi(1-y)^2 - pi(1-y^2)^2

my result is 7/6 pi..:( where am I wrong??
 
  • #4
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integrate it with respect to x. The inner radius is [tex] 1-\sqrt{x} [/tex] and the outer radius is [tex] 1- x [/tex].

So you have [tex] \int_{0}^{1} \pi((1-x)^{2}-(1-\sqrt{x})^{2}) \; dx [/tex]
 
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  • #5
arildno
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Well, your integrand is utterly wrong!
You are to rotate the VERTICAL strip [itex]\sqrt{x}\leq{y}\leq{x}[/itex] around y=1.
That gives you the integral:
[tex]V=\int_{0}^{1}\pi((1-x)^{2}-(1-\sqrt{x})^{2})dx=\pi\int_{0}^{1}(x^{2}-3x+2\sqrt{x})dx[/tex]
 
  • #6
arildno
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Apart from being wrongly signed, you have rotated your figure around the line x=1, rather than around y=1.
That accounts for your discrepancy.
(Note, for example, that your INNER radius is the actual OUTER radius)
 

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