Page 2 of 2 FirstFirst 12
Results 16 to 22 of 22

Math Help - Calculus questions

  1. #16
    Junior Member
    Joined
    Aug 2006
    Posts
    26
    Quote Originally Posted by ThePerfectHacker
    No,
    |x|\geq |y|
    |x|\leq 1
    ok, yep i see.

    so then if we split the integration into 2 regions, can't we just write this as the answer for part (b)?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by sarahisme
    ok, yep i see.

    so then if we split the integration into 2 regions, can't we just write this as the answer for part (b)?
    You are using symettry, good. But there is one thing which you might not have considered. The regions are symettrical, yes. But what about the surface above it! It happens to be true since,
    f(\sqrt{x^2+y^2} if replaced by -x,-y becomes, f(\sqrt{(-x)^2+(-y)^2}=f(\sqrt{x^2+y^2} so it is symettrical in the xz plane. Thus, that is valid.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Junior Member
    Joined
    Aug 2006
    Posts
    26
    Quote Originally Posted by ThePerfectHacker
    You are using symettry, good. But there is one thing which you might not have considered. The regions are symettrical, yes. But what about the surface above it! It happens to be true since,
    f(\sqrt{x^2+y^2} if replaced by -x,-y becomes, f(\sqrt{(-x)^2+(-y)^2}=f(\sqrt{x^2+y^2} so it is symettrical in the xz plane. Thus, that is valid.
    ok, sweet. if the surface above wasn't symmetric, would the answer be this for (b) :




    as for part (c), i am still struggling with it,

    for part (c) i get a graph like this (which is y = +/- sqrt(x-x^2)


    and so we want to integrate

    0<=x<=1
    0<=y<=sqrt(x-x^2)

    but when i try to convert to polar coordinates i end up in a mess....

    i think thats almost right? :S
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by sarahisme
    ok, sweet. if the surface above wasn't symmetric, would the answer be this for (b) :

    I assume so.

    (I am just cautioning you, I do not know all the specific rules for manipulating polar coordinates, I might be wrong). For example, when radius can be negative, when it cannot, .... Hate multivarible calculus, so informal )

    I do not understand your second problem, rephrase.
    as for part (c), i am still struggling with it,

    for part (c) i get a graph like this (which is y = +/- sqrt(x-x^2)


    and so we want to integrate

    0<=x<=1
    0<=y<=sqrt(x-x^2)

    but when i try to convert to polar coordinates i end up in a mess....

    i think thats almost right? :S[/QUOTE]
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Junior Member
    Joined
    Aug 2006
    Posts
    26
    ok rephrase, umm... i drew this graph

    which is a graph of y = +/- (x-x^2)^(1/2)



    now i want to integrate over that circle, right? so then i want to integrate over

    0<=x<=1
    0<=y<=sqrt(x-x^2)

    then i want to convert to polar coordinates.

    coverting 0<=x<=1 is easy enough, it just gives you 0<=r<=1/cos(theta)

    but i can't seem to convert 0<=y<=sqrt(x-x^2) to polar coordinates
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by sarahisme

    but when i try to convert to polar coordinates i end up in a mess....
    I understand you now.
    The most efficient way to solve this problem is with a Jacobain substitution. Since the circle is off center you can place it into the center (origon) by a basic substition of x for u+1, that will move the coordinate axes to the left 1 unit and substitute y for v because you are not changing anything.

    Doing it your way.
    x^2+y^2=x
    Thus,
    r^2=r\cos \theta
    Thus,
    r=\cos \theta
    Finished. I looks rather simple in polar form.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Junior Member
    Joined
    Aug 2006
    Posts
    26
    ok i think i almost have it now....

    for part (b) i get:



    for (c) i get this answer:

    \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{cos \theta}}f(tan\theta)rdrd\theta
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. T/F Calculus Questions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 12th 2010, 08:00 PM
  2. BC Calculus Questions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 13th 2009, 06:25 PM
  3. Calculus questions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 4th 2008, 11:46 AM
  4. Calculus [2 questions]
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 20th 2008, 07:56 AM
  5. pre-calculus questions
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: January 26th 2008, 08:28 PM

Search Tags


/mathhelpforum @mathhelpforum