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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.

enter image description here

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What's meant by $x=4$ about $x=1$? – Gigili May 30 '12 at 7:34
@Gigili Please check the picture – Radwa Kamal May 30 '12 at 7:51
The area bounded by : y^2=x^3 , x=4 about the axis X=1 – Radwa Kamal May 30 '12 at 8:19
when i put the picture I accidentally made mistakes but now it is correct – Radwa Kamal May 30 '12 at 8:23
@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? – mixedmath May 30 '12 at 8:23
up vote 0 down vote accepted

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 $-1<y<1$, 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 you – Radwa Kamal May 30 '12 at 16:01

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