Rotate the region bounded by x=y^2-6y+10 and x=5x about the y-axis.
Volume of body formed after rotation is-
V=∫152πyxdy=∫152πy(y2−6y+10)dyV=\int_1^52\pi yxdy=\int_1^52\pi y(y^2-6y+10)dyV=∫152πyxdy=∫152πy(y2−6y+10)dy
=2π[y44−2y3+5y2]15=2\pi [\dfrac{y^4}{4}-2y^3+5y^2]_1^5=2π[4y4−2y3+5y2]15
=2π[(6254−250+125)−(14−2+5)]=2\pi [(\dfrac{625}{4}-250+125)-(\dfrac{1}{4}-2+5)]=2π[(4625−250+125)−(41−2+5)]
=2π[1254−134]=2\pi[\dfrac{125}{4}-\dfrac{13}{4}]=2π[4125−413]
=112π2=56π cubic units =\dfrac{112\pi}{2}=56\pi \text{ cubic units }=2112π=56π cubic units
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments