Question #176671

Rotate the region bounded by x=y^2-6y+10 and x=5x about the y-axis.


1
Expert's answer
2021-04-15T17:11:23-0400

Volume of body formed after rotation is-


V=152πyxdy=152πy(y26y+10)dyV=\int_1^52\pi yxdy=\int_1^52\pi y(y^2-6y+10)dy


=2π[y442y3+5y2]15=2\pi [\dfrac{y^4}{4}-2y^3+5y^2]_1^5


  =2π[(6254250+125)(142+5)]=2\pi [(\dfrac{625}{4}-250+125)-(\dfrac{1}{4}-2+5)]


=2π[1254134]=2\pi[\dfrac{125}{4}-\dfrac{13}{4}]


  =112π2=56π cubic units =\dfrac{112\pi}{2}=56\pi \text{ cubic units }


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