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The function $x^2 = y\quad$ limits two areas $A$ and $B$:

$A$ is further limited with the line $x= a$, $a\gt 0$. $A$ rotates around the $x$-axis, which gives Volume $A = Va$.

$B$ is limited with the line $y=b$, $b\gt 0$. $B$ rotates around the $y$-axis, which gives Volume $B = Vb$.

What are the relations between $a$ and $b$, when $Vb = Va$?


I have come to the solution that:

$Vb = (\pi b^2 )/ 2$

$Va = (\pi a^5) / 5$

so the relation between them is:

$2.5b^2 = a^5$

Is that the final solution or is it more?

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If your calculations are correct this is what you should have found.

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Its homework right? – Tau May 21 '11 at 20:10
I was thinking maybe a simplification of my answer? – aka May 21 '11 at 20:29
No i've got my final exam on monday, so was thinking of training on old exams and other diverse stuff. – aka May 21 '11 at 20:31
Unless stated in the question that you have to produce some special answer, your answer should be excepted. Good luck on your test. – Tau May 21 '11 at 22:01
No it just says, what are the relations between a and b. when Va=Vb! Thank you Tau! – aka May 22 '11 at 14:41

Without a drawing or a more detailed description, I cannot be certain. But under the reasonable interpretation of what you wrote, your conclusion is absolutely correct. Maybe, since $a$ and $b$ are positive, it might be slightly better to say that $$b=a^2\sqrt{\frac{2a}{5}}$$

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Who did you get your solution? – aka May 21 '11 at 20:29
@aka: I am not sure of the meaning of your question. I drew the parabola, decided what the regions $A$ and $B$ were supposed to be, calculated the volumes of the solids, in each case by slicing, so for $A$ I integrated with respect to $x$ and for $B$ with respect to $y$. The integrations were trivial, as you discovered. – André Nicolas May 21 '11 at 20:47
The solution given above by user6312 is a simplification of your solution, in terms of expressing the relation b as a function of a: multiply both sides of your solution by 2/5 (the reciprocal of 2.5 = 5/2), then take the square root of both sides. – amWhy May 21 '11 at 20:50
I'm sorry i mean, how* ! – aka May 22 '11 at 14:31
@aka: I tried to describe it briefly in a comment above. I drew the diagram, and then calculated as explained in the very detailed explanation by @Arturo Magidin. I quickly obtained the expressions you wrote down, so took it for granted that you had done it the same way, so that a detailed working out of the details was not needed by you. Wrote down an alternate formula expressing $b$ in terms of $a$, in case you met this sort of question in a multiple choice test. – André Nicolas May 22 '11 at 15:05

If we take the region bounded by the $y$-axis, the $x$-axis, the line $x=a$ (with $a\gt 0$), and the parabola $y=x^2$, and rotate it about the $x$-axis, the volume of the resulting solid of revolution is easily computed (using, for example, discs perpendicular to the $x$-axis) to be $$\text{Volume A} = \int_0^a \pi(x^2)^2\,dx = \frac{\pi}{5}x^5\Bigm|_0^a = \frac{\pi a^5}{5}.$$ If the region bounded by the $y$-axis, the $x$ axis, the line $y=b$ (with $b\gt 0$), and the parabola $y=x^2$ is revolved around the $y$-axis, then using discs perpendicular to the $y$-axis we obtain the volume to be: $$\text{Volume B} = \int_0^b \pi (\sqrt{y})^2\,dy = \frac{\pi}{2}y^2\Bigm|_0^b = \frac{\pi b^2}{2}.$$ So your computations are correct there.

If the two volumes are the same, then we must have $$\text{Volume A} = \frac{\pi a^5}{5} = \frac{\pi b^2}{2} = \text{Volume B};$$ there are many ways to express this: you can solve for one of $a$ or $b$ in terms of the other: $$b = \sqrt{\frac{2a^5}{5}} = a^{5/2}\sqrt{\frac{2}{5}},$$ or, if you want to express $a$ in terms of $b$ instead, $$ a = \sqrt[5]{\frac{5}{2}b^2} = b^{2/5}\sqrt[5]{\frac{5}{2}}.$$ Or you can simply express this relation by saying, say $$2a^5 = 5b^2.$$

Note. If $a\lt 0$, then the volume of $A$ can be computed the same way, but the integral would go from $a$ to $0$, so that the volume would be $-\frac{\pi a^5}{5}$; to account for both possibilities, both $a\gt 0$ and $a\lt 0$, you can simply write that the volume is $\frac{\pi|a|^5}{5}$. For solid $B$, however, it makes no sense to talk about $b\lt 0$, because then we don't have a finite area "enclosed" by the curves in question.

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I did exacly the same calculations, was just a bit confused by the simplification you made in the last steps... but i think it is as you say... 2a^5 = 5b^2 2a^5/5 = b^2 then take square root we get: sqroot ((2a^5)/5) = b b = a^2* sqroot(2a/5) if we want a: a^5 = (5b^2/2) a = 5root(5b^2/2) a = b^(2/5) * 5root(5/2) – aka May 22 '11 at 14:39

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