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## Main Question or Discussion Point

**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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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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arildno

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Pinpoint your problem.

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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??

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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]

So you have [tex] \int_{0}^{1} \pi((1-x)^{2}-(1-\sqrt{x})^{2}) \; dx [/tex]

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arildno

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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]

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arildno

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That accounts for your discrepancy.

(Note, for example, that your INNER radius is the actual OUTER radius)

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