1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig Substitution and Volume of Solids

  1. Nov 9, 2011 #1
    2 Questions here! (I'm not exactly sure if it's allowed, but I want to avoid posting too many threads)

    1. The problem statement, all variables and given/known data
    1)Find the following definite integrals by using a trigonometric substitution:
    d)1/21dx/(√(2x-x2)


    2. Relevant equations
    x=asinu


    3. The attempt at a solution
    From a previous question, I found that the indefinite integral of ∫dx/(√(2x-x2) was arcsin(x-1) + C
    Using FTC part 2, I got:
    arcsin(1-1)-arcsin(1/2-1)=arcsin0-arcsin(-1/2)=0- -∏/6=∏/6

    The answer the teacher gave me was ∏/2
    I'm have no idea where I have made a mistake.


    1. The problem statement, all variables and given/known data
    Find the volumes of the solids generated by revolving the region bounded by the curves y=x+2 and y=x2 about (a) the line x=2 (b) the line y=4


    2. Relevant equations
    Shell:V= 2∏abxhdx
    Disc: V=∏abr22-r12dx


    3. The attempt at a solution
    a)
    x+2=x2
    0=x2-x+2
    (-b±√(b2-4ac))/2a=(1±√(1-4(1)(-2))/2(1)=(1±3)/2
    x=-1 and x=2

    2∏-12x*(x+2-x2)dx
    2∏(1/3*x3+x2-1/4x4|-12
    2∏(8/3-5/12)=

    9∏/2
    The correct answer was supposed to be 27∏/2


    b)
    ab(4-(x2)2(4-(x+2))2dx
    ab12-9x2+4x-x4dx
    ∏(12x-3x3+2x[sup2]-1/5x[sup5])|-12
    ∏[(24-24+8-32/5)-(-12+3+2+1/5)]=
    42∏/5
    The answer is supposed to be 108∏/5


    Thanks!
     
  2. jcsd
  3. Nov 9, 2011 #2
    For d) It is not
    [tex]arcsin(x-1)[/tex]
     
  4. Nov 9, 2011 #3

    I like Serena

    User Avatar
    Homework Helper

    Welcome to PF, Luscinia! :smile:

    No mistake, your answer to (d) is correct.
    It is arcsin(x-1) + C and it evaluates to ∏/6.
    (Sorry sandy.bridge.)
     
  5. Nov 9, 2011 #4
    Oops, you're right. I presume that your issue was with regards to your lower and upper limits. Did you remember to alter them during substitution?
     
  6. Nov 9, 2011 #5
    Thanks for the quick answer!
    I'm not sure why I need to replace the upper and lower limit since the final integral for ∫dx/(√(2x-x2) uses the variable x as well.
    (Though I honestly forgot about them and I'm not too sure if I know how to do it properly)
    I had used (x-1)=sinu as substitution when I had integrated so arcsin(x-1)=arcsin(sinu)=u

    (1/2-1)=sinu
    arcsin(-1/2)=u=-∏/6 (lower limit)

    (1-1)sinu
    arcsin(0)=u=0 (upper limit)

    using FTC:
    0--∏/6=∏/6?

    Maybe my teacher made a mistake?
     
  7. Nov 9, 2011 #6
    Okay, I just did the question and got the same answer as you. Your teacher must have given you a false answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trig Substitution and Volume of Solids
  1. Trig substitution (Replies: 5)

  2. Trig substitution (Replies: 4)

  3. Trig substitution (Replies: 7)

  4. Trig Substitution (Replies: 2)

Loading...