What is the relation?

The function $x^2 = y\quad$ restricts 2 locations $A$ and also $B$:

$A$ is better restricted with the line $x= a$, $a\gt 0$. $A$ revolves around the $x$ - axis, which offers Volume $A = Va$.

$B$ is restricted with the line $y=b$, $b\gt 0$. $B$ revolves around the $y$ - axis, which offers Volume $B = Vb$.

What are the relationships in between $a$ and also $b$, when $Vb = Va$?

I have involved the remedy that:

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

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

so the relationship in between them is:

$2.5b^2 = a^5$

Is that the last remedy or is it extra?

2022-06-07 14:33:27
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Answers: 3

If your estimations are proper this is what you need to have located.

2022-06-07 14:52:49

Without an illustration or an extra thorough summary, I can not be particular. Yet under the practical analysis of what you created, your verdict is definitely proper. Possibly, given that $a$ and also $b$ declare, it could be a little far better to claim that $$b=a^2\sqrt{\frac{2a}{5}}$$

2022-06-07 14:52:26

If we take the area bounded by the $y$ - axis, the $x$ - axis, the line $x=a$ (with $a\gt 0$), and also the parabola $y=x^2$, and also revolve it concerning the $x$ - axis, the quantity of the resulting strong of change is conveniently calculated (making use of, as an example, discs vertical 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 area bounded by the $y$ - axis, the $x$ axis, the line $y=b$ (with $b\gt 0$), and also the parabola $y=x^2$ is focused on the $y$ - axis, after that making use of discs vertical to the $y$ - axis we get the quantity 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 calculations are proper there.

If both quantities coincide, after that we have to have $$\text{Volume A} = \frac{\pi a^5}{5} = \frac{\pi b^2}{2} = \text{Volume B};$$ there are several means to share this: you can address for among $a$ or $b$ in regards to the various other: $$b = \sqrt{\frac{2a^5}{5}} = a^{5/2}\sqrt{\frac{2}{5}},$$ or, if you intend to share $a$ in regards to $b$ rather, $$ a = \sqrt[5]{\frac{5}{2}b^2} = b^{2/5}\sqrt[5]{\frac{5}{2}}.$$ Or you can merely share this relationship by claiming, claim $$2a^5 = 5b^2.$$

Note. If $a\lt 0$, after that the quantity of $A$ can be calculated similarly, yet the indispensable would certainly go from $a$ to $0$, to make sure that the quantity would certainly be $-\frac{\pi a^5}{5}$ ; to make up both opportunities, both $a\gt 0$ and also $a\lt 0$, you can merely write that the quantity is $\frac{\pi|a|^5}{5}$. For strong $B$, nonetheless, it makes no feeling to speak about $b\lt 0$, due to the fact that after that we do not have a limited location "enclosed" by the contours concerned.

2022-06-07 14:52:14