Answer to Question #176671 in Calculus for Joshua

Question

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

Answer 1

Volume of body formed after rotation is-

$V=\int_1^52\pi yxdy=\int_1^52\pi y(y^2-6y+10)dy$

$=2\pi [\dfrac{y^4}{4}-2y^3+5y^2]_1^5$

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

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

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