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Question Number 136076 by physicstutes last updated on 18/Mar/21

$$\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{between}\mathrm{the}\mathrm{curve}y=3+2x-x^2,\mathrm{the}x-\mathrm{axis}$$$$\mathrm{and}\mathrm{the}\mathrm{line}y=3.$$$$\mathrm{find}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{generated}\mathrm{when}\mathrm{the}\mathrm{curve}\mathrm{is}$$$$\mathrm{rotated}\mathrm{completely}\mathrm{about}\mathrm{the}\mathrm{line}y=3$$

Answered by mr W last updated on 18/Mar/21

$$AreaA=\frac{2}{3}\times (4\times 4-2\times 1)=\frac{28}{3}$$

Commented by mr W last updated on 18/Mar/21

Commented by mr W last updated on 18/Mar/21

$$volumeV$$$$x^2-2x-3+y=0$$$${(x_2-x_1)}^2={(x_2+x_1)}^2-4x_1{x}_2$$$$=2^2-4(-3+y)=4(4-y)$$$$x_2-x_1=2\sqrt{4-y}$$$$dV=2\pi (3-y)(x_2-x_1)dy$$$$V={\int }_0^{3}2\pi (3-y)(x_2-x_1)dy$$$$=4\pi {\int }_0^3(3-y)\sqrt{4-y}dy$$$$=4\pi {\int }_3^0(3-y)\sqrt{1+3-y}d(3-y)$$$$=4\pi {\int }_0^{3}u\sqrt{1+u}du$$$$=4\pi \times \frac{2}{15}{\left[{(u+1)}^{\frac{3}{2}}(3u-2)\right]}_0^3$$$$=4\pi \times \frac{2\times 58}{15}$$$$=\frac{464\pi }{15}$$

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