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Homework Help: Integral of hell need help. alot of work is shown

  1. Mar 9, 2009 #1
    Evaluate the integral below:

    2Pi times the integral of (2-y^2)(sqrt(1+4y^2)) dy from 0 to sqrt(2)



    work shown:

    using integration by parts I got:

    let u= sqrt(1+4y^2)

    therefore du= 4y/(4y^2+1)

    let dv= (2-y^2)

    therefore V= 2y- y^3/3

    since Integration by Parts is (u)(V) - integral of (V)(du)

    therefore i got 2*Pi times (sqrt(1+4y^2)(2y-y^3/3) - integral of (2y-y^3/3)(4y/sqrt(4y^2+1)) dy from 0 to sqrt (2)


    using trig. subs. method i got

    i tried using the tan y= 2y method of trigonometric substition

    but i still wasn't able to evaluate the integral fully

    i still need some help...

    here's my work shown below:

    2Pi times integral of (1-2y^2)*(sqrt(1+4y^2)) from 0 to sqrt(2) becomes:

    2PI times integral of (1- ytan y)(sqrt(1+tan^2(y)) from 0 to sqrt(2)

    since sec^2(y)= 1+tan^2(y)

    i got 2Pi times integral of (1- ytan(y))(sqrt(sec^2(y))

    therefore i got 2Pi times integral of (1- y*tan(y))(sec(y))

    since sec y= 1/ cos y and tan x= sin y/ cos y

    if i let u = sec y
    du= sec y tan y dy

    therefore i got, 2Pi times integral of u - y du from 0 to sqrt(2)

    i still need a lot of help... i've tried using normal integration by parts methods.. and i haven't been able to fully evaluate the integrate because it keeps creating another integral that needs to evaluated by parts again.. .i need a lot of help here
    after evaluating this i keep getting a more completed integration by parts.. and continous iteration.. please help me solve this integral

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 10, 2009 #2

    Mark44

    Staff: Mentor

    The first thing you should do is break this into two integrals.
    [tex]2\pi \int_0^{\sqrt{2}} (2 - y^2)\sqrt{1 + 4y^2}dy[/tex]
    [tex]= 2\pi \int_0^{\sqrt{2}} 2\sqrt{1 + 4y^2}dy - 2\pi \int_0^{\sqrt{2}} y^2 \sqrt{1 + 4y^2}dy[/tex]

    I think that both can be tackled with the same trig substitution, namely tan u = 2y.
     
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