The volume of the solid generated by revolvin | vedclass

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Question

(C) The curves are $y=x^{2}$ (or $x=\sqrt{y}$) and $x=y^{2}$.They intersect at $(0,0)$ and $(1,1)$.When revolving about the $y$-axis,the volume $V$ is

Vedclass · Accepted Answer

$\frac{3}{10} \pi$

Vedclass · Answer

(C) The curves are $y=x^{2}$ (or $x=\sqrt{y}$) and $x=y^{2}$.They intersect at $(0,0)$ and $(1,1)$.When revolving about the $y$-axis,the volume $V$ is given by the formula $V = \pi \int_{a}^{b} (x_{outer}^{2} - x_{inner}^{2}) dy$.Here,for $y \in [0,1]$,the outer curve is $x = \sqrt{y}$ and the inner curve is $x = y^{2}$.Thus,$V = \pi \int_{0}^{1} ((\sqrt{y})^{2} - (y^{2})^{2}) dy$$V = \pi \int_{0}^{1} (y - y^{4}) dy$$V = \pi \left[ \frac{y^{2}}{2} - \frac{y^{5}}{5} \right]_{0}^{1}$$V = \pi \left( \frac{1}{2} - \frac{1}{5} \right) = \pi \left( \frac{5-2}{10} \right) = \frac{3}{10} \pi$.

The volume of the solid generated by revolving the region bounded by the parabolas $y=x^{2}$ and $x=y^{2}$ about the $y$-axis is:

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