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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 $Vb = Va$?

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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?

3 Answers 3

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

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    No it just says, what are the relations between a and b. when Va=Vb! Thank you Tau!2011-05-22
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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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    I wish it was$a$multiple choice test... they want calculations! But many thanks for your help Both User6312 and Arturo!2011-05-22
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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)2011-05-22