Results 1 to 6 of 6

Math Help - volume of revolutions

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    91

    volume of revolutions

    find the volume of the solid generated by revolving about the y-axis the region bounded by x^2 + y^2 = a^2 , y = b>0 , x = 0
    for b<=y<=a
    (a spherical cap)

    i assumed that y=b>0 meant to only count the graph above y=b
    the answer i got was:

    Pi * [ (a^4 -(a^6 / 3)) - ( (a^2)*b - (b^3 / 3) ) ]

    sorry about all the brackets , i tried to make it as clean as possible

    can someone check whether this is the correct answer and if not point me in the right direction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by razorfever View Post
    find the volume of the solid generated by revolving about the y-axis the region bounded by x^2 + y^2 = a^2 , y = b>0 , x = 0
    for b<=y<=a
    (a spherical cap)

    i assumed that y=b>0 meant to only count the graph above y=b
    the answer i got was:
    Pi * [ (a^4 -(a^6 / 3)) - ( (a^2)*b - (b^3 / 3) ) ]

    sorry about all the brackets , i tried to make it as clean as possible

    can someone check whether this is the correct answer and if not point me in the right direction
    so assuming you drew the graph correctly, you will notice that the shell method seems like the more straightforward way to go here.

    the integral we must compute is:

    V = 2 \pi \int rh~dx

    = 2 \pi \int_0^{\sqrt{a^2 - b^2}} x(\sqrt{a^2 - x^2} - b)~dx

    = 2 \pi \int_0^{\sqrt{a^2 - b^2}} x \sqrt{a^2 - x^2}~dx - 2 \pi \int_0^{ \sqrt{a^2 - b^2}} bx~dx

    the second integral is easy, for the first, a substiutiton of u = a^2 - x^2 will work nicely. or even u^2 = a^2 - x^2 if you're up for it
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    91
    how does the graph look like
    can you describe it in words
    or post a link to it or something
    i really want to understand the question ... not just solve it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by razorfever View Post
    how does the graph look like
    can you describe it in words
    or post a link to it or something
    i really want to understand the question ... not just solve it
    here is the graph (see below). the shaded region is what you are rotating about the y-axis. i made the radius of the circle you see and the position of the horizontal line arbitrary, so don't mind the values that you see, just note what the general shape will look like.

    now, do you remember how the shell method says to proceed?
    Attached Thumbnails Attached Thumbnails volume of revolutions-graph.png  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    Posts
    91
    by solving the integral

    <br /> <br />
= 2 \pi \int_0^{\sqrt{a^2 - b^2}} x \sqrt{a^2 - x^2}~dx - 2 \pi \int_0^{ \sqrt{a^2 - b^2}} bx~dx<br />

    the answer i get is:

    2Pi * ( (2/3)a^3 * (-sqrt(a^2-b^2) - 1))

    is this the correct answer?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by razorfever View Post
    find the volume of the solid generated by revolving about the y-axis the region bounded by x^2 + y^2 = a^2 , y = b>0 , x = 0
    for b<=y<=a
    (a spherical cap)

    i assumed that y=b>0 meant to only count the graph above y=b
    the answer i got was:

    Pi * [ (a^4 -(a^6 / 3)) - ( (a^2)*b - (b^3 / 3) ) ]

    sorry about all the brackets , i tried to make it as clean as possible

    can someone check whether this is the correct answer and if not point me in the right direction
    Duplicate post.

    Things will be too messy trying to move the replies to here http://www.mathhelpforum.com/math-he...ut-y-axis.html and then edit so that it all makes sense.

    Easier just to close this thread.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Car Wheel Revolutions?
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: August 29th 2010, 12:19 PM
  2. Volume of solids of revolutions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 14th 2010, 01:53 AM
  3. Revolutions help, please!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 30th 2008, 02:55 PM
  4. Revolutions Per Minutes
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 1st 2008, 04:03 PM
  5. revolutions per minute??
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 19th 2007, 08:45 PM

Search Tags


/mathhelpforum @mathhelpforum