Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    find the area of the surface obtained by revolving the curve y=x^2, 0<x<1 about the y axis
    Offline

    13
    ReputationRep:
    use the formula attached:
    Attached Images
     
    Offline

    10
    ReputationRep:
    y = x^2 => x = y^1/2

    x = 0, y = 0, x = 1, y = 1

    Surface area, S = 2*pi*integral [0, 1] of g(y)(1 + [g'(y)]^2)^1/2

    g(y) = y^1/2
    g'(y) = (1/2)y^-1/2

    [0,1] are the limits.

    You should be able to do it from there.
    Offline

    10
    ReputationRep:
    (Original post by elpaw)
    use the formula attached:
    Damn you elpaw!
    • Thread Starter
    Offline

    0
    ReputationRep:
    thanx u guys
    Offline

    13
    ReputationRep:
    (Original post by Simbo)
    thanx u guys
    sorry, i didnt see te "about y axis". The formula for your case is thus....
    Attached Images
     
 
 
 
Poll
“Yanny” or “Laurel”

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.