Asked by Melissa
Find the area of the surface generated when y=4x and x=1 is revolved about the y-axis.
Answers
Answered by
Steve
No calculus need for this one. It's just a cylinder with a cone cut out.
r=1
h=4
v = πr^2h - π/3 r^2h = 2π/3 r^2h = 8π/3
Ok ok. If you want to use calculus, then with shells,
v = ∫[0,1] 2πrh dx
where r=x and h=y=4x
v = ∫[0,1] 2πx*4x dx = 8π/3
using discs (washers) you get
v = ∫[0,4] π(R^2-r^2) dy
where R=1 and r=x=y/4
∫[0,4] π(1-y^2/16) dy = 8π/3
r=1
h=4
v = πr^2h - π/3 r^2h = 2π/3 r^2h = 8π/3
Ok ok. If you want to use calculus, then with shells,
v = ∫[0,1] 2πrh dx
where r=x and h=y=4x
v = ∫[0,1] 2πx*4x dx = 8π/3
using discs (washers) you get
v = ∫[0,4] π(R^2-r^2) dy
where R=1 and r=x=y/4
∫[0,4] π(1-y^2/16) dy = 8π/3
Answered by
Melissa
We have to use the Surface area of revolution formula
integral(f(x) aqrt(1+f'(x)^2))
integral(f(x) aqrt(1+f'(x)^2))
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.