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Question Number 194831 by necx122 last updated on 16/Jul/23

Commented by necx122 last updated on 16/Jul/23

$$I^{\prime }vetriedattemptingthisquestionbut$$$$gotstuck.Please,Ineedassistance.$$

Answered by dimentri last updated on 16/Jul/23

$$V=\pi \underset{0}{\stackrel{\pi }{\int }}\sin {}^{2}xdx$$$$V=\pi {[\frac{1}{2}x-\frac{1}{4}\sin 2x]}_0^{\pi }$$$$V=\frac{1}{2}{\pi }^2$$

Commented by necx122 last updated on 16/Jul/23

$$Thankyousir.$$

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