The part of the circle x^2 + y^2 = 9 between | vedclass

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(A) The part of the circle $x^2 + y^2 = 9$ between $y = 0$ and $y = 2$ is revolved about the $y$-axis.This generates a solid of revolution.The volume

Vedclass · Accepted Answer

$\frac{46}{3}\pi$

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(A) The part of the circle $x^2 + y^2 = 9$ between $y = 0$ and $y = 2$ is revolved about the $y$-axis.This generates a solid of revolution.The volume $V$ is given by the formula $V = \pi \int_{a}^{b} x^2 \, dy$.From the equation of the circle,$x^2 = 9 - y^2$.Substituting the limits $y = 0$ to $y = 2$:$V = \pi \int_{0}^{2} (9 - y^2) \, dy$$V = \pi \left[ 9y - \frac{y^3}{3} \right]_{0}^{2}$$V = \pi \left[ (9(2) - \frac{2^3}{3}) - (0 - 0) \right]$$V = \pi \left[ 18 - \frac{8}{3} \right]$$V = \pi \left[ \frac{54 - 8}{3} \right] = \frac{46}{3}\pi$ cubic units.

The part of the circle $x^2 + y^2 = 9$ between $y = 0$ and $y = 2$ is revolved about the $y$-axis. The volume of the generating solid will be:

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