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Question Number 169756 by mokys last updated on 07/May/22

Answered by aleks041103 last updated on 07/May/22

$$A.y=\frac{3}{x}$$$$V={\int }_2^3\pi y^{2}dx=9\pi {\int }_2^3\frac{dx}{x^2}=9\pi {\left(\frac{1}{x}\right)}_3^2=$$$$=9\pi (\frac{1}{2}-\frac{1}{3})=\frac{9\pi }{6}=\frac{3\pi }{2}=V$$

Answered by aleks041103 last updated on 07/May/22

$$B.$$$$\left(a\right)V={\int }_0^3\pi y^{2}dx=\pi {\int }_0^3{(2x^2+3)}^{2}dx=$$$$=\pi {\int }_0^3(4x^4+12x^2+9)dx=$$$$=\pi {[\frac{4}{5}{x}^5+4x^3+9x]}_0^3=$$$$=\pi (\frac{4.3^5}{5}+4.3^3+9.3)=$$$$=\pi (\frac{36.3^3}{5}+5.3^3)=$$$$=\frac{27\pi }{5}(36+25)=\frac{27.61\pi }{5}$$

Answered by thfchristopher last updated on 08/May/22

$$\mathrm{A}.$$$$\mathrm{For}\mathrm{the}\mathrm{curve}\mathrm{rotates}\mathrm{x}-\mathrm{axis},$$$$y=\frac{3}{x}$$$$V=\pi {\int }_2^3{\left(\frac{3}{x}\right)}^{2}dx$$$$=\pi {\int }_2^3\frac{9}{x^2}dx$$$$=-\pi {\left[\frac{9}{x}\right]}_2^3$$$$=\frac{3\pi }{2}$$$$$$$$\mathrm{B}.$$$$\left(\mathrm{a}\right)\mathrm{For}\mathrm{the}\mathrm{curve}\mathrm{rotates}\mathrm{x}-\mathrm{axis},$$$$V_x=\pi {\int }_0^3{(2x^2+3)}^{2}dx$$$$=\pi {\int }_0^3(4x^4+12x^2+9)dx$$$$=\pi {[\frac{4x^5}{5}+4x^3+9x]}_0^3$$$$=\frac{783\pi }{5}$$$$\left(\mathrm{b}\right)\mathrm{For}\mathrm{the}\mathrm{curve}\mathrm{rotates}\mathrm{y}-\mathrm{axis},$$$$x=\sqrt{\frac{y-3}{2}}$$$$\mathrm{When}x=3,y=21$$$$x=0,y=3$$$$\therefore V_y=\pi {\int }_3^{21}\frac{y-3}{2}dy$$$$=\pi {[\frac{y^2}{4}-\frac{3y}{2}]}_3^{21}$$$$=81\pi$$

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