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Find the volume generated by revolving the area bounded by $y^2=x^3$ , $x=4$ about the line $x=1$.

I can't understand how the area will revolve about a line lying on the area. Many thanks in advance.

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    What's meant by $x=4$ about $x=1$?2012-05-30
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    @Gigili Please check the picture2012-05-30
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    The area bounded by : y^2=x^3 , x=4 about the axis X=12012-05-30
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    when i put the picture I accidentally made mistakes but now it is correct2012-05-30
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    @Phony: one would assume that you have 'implied region of integration' $[1,4]$. At least, that's what I would assume. Can you do it now?2012-05-30
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    my problem I can't get how can I revolve a region about an axis lies within the region ?? how can this be possible ?2012-05-30
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    @PhonyPearl: Sorry about the edit, I thought you meant the interval.2012-05-30
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    @Phony: That's why I said you probably need to use the region $[1,4]$, so that the axis is not inside the region, so to speak.2012-05-30
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    I think you are given an area with the axis going through it, so what happens when you revolve the figure - can you reconfigure the problem as rotating one or more areas without an axis through the middle?2012-05-30
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    @mixedmath how about the other region [0.1] is n't it suppose to revolve also or I take the larger area as the smaller will be included during revolution of the larger area??2012-05-30
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    @MarkBennet I can't get what you mean ??2012-05-30
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    One of the things you learn about solving problems is that you are sometimes given a problem in a form which looks unfamiliar, and to solve it, you sometimes need to wrestle it into a form you know how to solve. I would suggest sketching the solid of rotation, not just the graph - and then see how the graph relates to the volume you are supposed to compute. Then you might see why mixed math suggests you can ignore the part of the graph which is on one side of the axis of rotation.2012-05-30
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    @MarkBennet what if I worked on the other interval the result will differ right ??2012-05-30

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The volume of a solid of revolution is the integral of the square of the distance from the axis of revolution times $\frac{\tau}2=\pi$. The part of the curve to the left of $x=1$, where $-1, will be inside the solid formed as the rest of the curve rotates so can be ignored.

volume $=\frac{\tau}2 (\int_{1}^{8}( \ ^3\sqrt{y^2}-1)^2 \ dy+\int_{-8}^{1}( \ ^3\sqrt{y^2}-1)^2 \ dy)$

$\int (^3\sqrt{y^2}-1)^2dy=\int y^{\frac{4}3}-2y^{\frac{2}3}+1 \ dy=\frac{3}7 y^{\frac{7}3}-\frac{6}5 y^{\frac{5}3}+y$

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    Many Thanks for all of you2012-05-30