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!

Simple Integration

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    question.jpg


    3. The attempt at a solution

    there are a number of things i've had trouble with here,

    the first part i got fine, bieng;

    y^2 = 4x^2 ( 9 - x^2) , which is correct according to the answers, next

    the 'show that' question, now i had no problem getting it into the right form using the;

    [tex]\int f(t)\frac{dx}{dt}dt[/tex] rule, so i put it in like so;

    f(t) = 9 sin (2t) and [tex]\frac{dx}{dt}[/tex] = -3 sin t,

    so i put them together to get, [tex]\int -27 sin(2t) sin(t)[/tex] , so A, the constant, would be minus 27...wrong! it simply says 27 in the back, so i thought, hey well maybe it's just an error, but i have a sneaking suspicion i might be wrong aswell so i would appreciate clarification on that one aswell.

    next, "find the value of this integral".

    i don't know but this is a hard, hard integral, and we havn't even covered integration by parts or any formal integration aside from the [tex]\frac{n^(n+1)}{n+1}[/tex] stuff, I have however done "some" differential equations in my own time and am somewhat familiar with integration, i can wade my way through it sometimes however this one strikes me as very hard, i thought, well maybe i can break down the sin (2t), giving me,

    [tex]\int cos(t) - (cos(t))^3[/tex] however from here I am also stumped....

    also, it says to integrate between [tex]\frac{\pi}{2}[/tex] and 0, why [tex]\frac{\pi}{2}[/tex] i wonder? intuitiveley i looked at the equation and expanded the 9 - x^2 showing that the roots were x = 3 & x = -3, so if you put 3 (clearly) into the x equation to work out the value for t, 3 = 3 cos (t) -> 1 = cos (t) t = 0, 2[tex]/pi[tex] , so why [tex]\frac{\pi}{2}[/tex]?

    i would be much obliged if someone could give some advice on this one!

    however i did look up the answer for this so as to complete the next part of the question, which was 18, and was able to do the last question

    thanks guys i realise it's a long question but i think it's just me and it might only take someone good at maths 2 minutes :P, sorry again for the huge post and thank you
     
  2. jcsd
  3. Sep 7, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Well it looks to me that the -ve sign would say that you re finding the area of the opposite side that is shaded (the part under the x-axis)

    For the integral try using sin2t=2sintcost and then using a u substitution
     
  4. Sep 7, 2009 #3

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    For the second one, note that it is actually
    [tex]
    \int f(t) \left|\frac{dx}{dt}\right| dt.
    [/tex]
    Geometrically it also makes sense that A > 0: the integral has a geometrical meaning as surface, which is always positive.

    For the third question, you might find it helpful that
    [tex]\sin(2t) = 2 \sin(t) \cos(t)[/tex]
    and use a substitution of variables.

    (If the latter doesn't mean anything to you: try writing the integrand as a derivative f'(t) of some function f(t)).
     
  5. Sep 7, 2009 #4
    By compound angle formula, (sin 2t)(sin t) = 1/2 ( cos (t) - cos (3t) ). That's an easily integrable form
     
  6. Sep 7, 2009 #5
    hmm thanks alot for the clearing up on the 2nd and 3rd question however i'm still wondering about the integrating between

    pi/2 and 0 has come from, as on the x-axis you're integrating between 3 & 0, meaning 2pi and 0 for t....
     
  7. Sep 7, 2009 #6

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    For t = 0, the curve passes through (x, y) = (3, 0) and runs counter-clockwise. The upper integration limit is the time instant where it passes through the origin. When you solve for x = 0, y = 0, you will find that the first solution is t = pi / 2 (remember, you only need to integrate over the time interval in which the curve encloses the shaded area).
     
  8. Sep 7, 2009 #7
    x = 3 cos t adopts the range of 0 to 3 for t having a range of 0 to pi/2. While other ranges appear to satisfy this condition, we must ensure that y= 9 sin 2t must be positive in that range of t chosen as well.
     
  9. Sep 7, 2009 #8
    Ahh i see, so the 'origin' for t is actually (3,0) ? :OO

    is that a general rule for all parametric forms - finding out the (x,y) when it's 0 and integrating FROM that point?

    much obliged, this assistance is top quality
     
  10. Sep 7, 2009 #9

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    No. In principle, what you want to do is calculate the area under a part of the curve [itex]y^2 = f(x)[/itex]. You need to find out which part that is. When you plug in t = 0 in the equation, you get (x, y) = (3, 0), which is the point all the way on the right on the x-axis. When you look at the shaded area, you see that you have to integrate up to the point where the graph crosses the x-axis (which, in this case, happens to be the origin). So you need to find which t belongs to this point. I suggest you do the calculation yourself and check that it gives t = pi / 2.
     
  11. Sep 7, 2009 #10
    that's great, much obliged sir
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple Integration
  1. Simple integration (Replies: 2)

  2. Simple Integral (Replies: 3)

  3. Simple Integral (Replies: 10)

  4. Simple integral (Replies: 7)

  5. Simple Integral (Replies: 8)

Loading...