Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    The part of the curve y=x^2 between y=0 and y=2 is rotated completely about the y-axis. show that the area of the curved surface formed is 13pie/3


    I am ok if it goes round the x-axis but what happens when its round the y-axis? can anybody help me out please?
    Offline

    15
    ReputationRep:
    (Original post by Mr_Joe_T)
    The part of the curve y=x^2 between y=0 and y=2 is rotated completely about the y-axis. show that the area of the curved surface formed is 13pie/3


    I am ok if it goes round the x-axis but what happens when its round the y-axis? can anybody help me out please?
    SA=2pi \int_{y_{a}}^{y_{b}} x[1+(\frac{dx}{dy})^2]^{\frac{1}{2}} dy\\



x=y^{\frac{1}{2}}\\



\frac{dx}{dy} = \frac{1}{2}y^{\frac{-1}{2}}\\



(\frac{dx}{dy} )^{2} = \frac{1}{4}y^{-1}\\



1 + (\frac{dx}{dy} )^{2} = (\frac{1}{4y})(4y+1)\\



SA=2pi \int_0^2 (\frac{1}{2})[4y+1]^{\frac{1}{2}} dy\\



SA=(\frac{\pi}{6})[(4y+1)^{\frac{3}{2}}]_0^2\\



SA=(\frac{\pi}{6})[26]\\



SA=\frac{13\pi}{3}\\
    Offline

    0
    ReputationRep:
    (Original post by Gaz031)
    SA=2pi \int_{X_{a}}^{X_{b}} x[1+(\frac{dx}{dy})^2]^{\frac{1}{2}} dy
    If you need trouble evaluating the integral then post back.
    Shouldn't the limits Xa and Xb be Ya and Yb ?
    Or am i wrong?
    Offline

    0
    ReputationRep:
    ye they should be
    Offline

    0
    ReputationRep:
    (Original post by Syncman)
    ye they should be
    oh good....i just had a minor panic that i'd learnt it wrong!
    Offline

    0
    ReputationRep:
    How would you integrate that?
    Substitution of y=1/4 sinh²u?
    Offline

    2
    ReputationRep:
    I got dy/dx as 1/(2√y)

    then got it as 2pi∫(√y√((4y+1)/4y)
    =(2/3)pi(4y+1)^3/2

    which between limits gives an area of (18-2/3)pi=52pi/3


    what did i do wrong :/
    Offline

    15
    ReputationRep:
    (Original post by lgs98jonee)
    I got dy/dx as 1/(2√y)

    then got it as 2pi∫(√y√((4y+1)/4y)
    =(2/3)pi(4y+1)^3/2

    which between limits gives an area of (18-2/3)pi=52pi/3


    what did i do wrong :/
    I edited my post.
    Offline

    0
    ReputationRep:
    Has anyone done question 8 P5 June 2004? please help!! (edexcel)
    Offline

    0
    ReputationRep:
    (Original post by Lucky Penny)
    Has anyone done question 8 P5 June 2004? please help!! (edexcel)
    yeah. which part u stuck on?
    Offline

    8
    ReputationRep:
    (Original post by majikthise)
    How would you integrate that?
    Substitution of y=1/4 sinh²u?
    Nope!
    you end up with,

    Sa =2π ∫√y√(1+1/(4y) dy
    = 2π ∫ √(y + ¼) dy
    = 2π ∫ (y + ¼)^½ dy
    = 2π(2/3)(y + ¼)^(3/2) [ 0 -> 4]
    = 2π (2/3){(9/4)^(3/2) - (1/4)^(3/2)}
    = 2π (2/3){(3/2)^3 - (1/2)^3}
    = 2π (2/3){27/8 - 1/8}
    = (4π/3){26/8}
    = 13π/3
    =====
 
 
 
Turn on thread page Beta
Updated: June 19, 2005
The home of Results and Clearing

2,253

people online now

1,567,000

students helped last year

University open days

  1. SAE Institute
    Animation, Audio, Film, Games, Music, Business, Web Further education
    Thu, 16 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  3. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
Poll
Do you want your parents to be with you when you collect your A-level results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

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.