Results 1 to 2 of 2

Math Help - volumes rotate about y = x

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    volumes rotate about y = x

    i don't know how to start this problem. Instead of rotating about a horizontal and vertical line, it requires you to rotate about the line y =x

    determine the volume bounded by the curve when y = x^4 and y =x between x = 0 and x = 1 are rotated about the line y =x

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,792
    Thanks
    1532
    That's going to be a difficult problem but I think I would be inclined to rotate the coordinate system so as to make the line y= x correspond to the new y-axis. That is, we must rotate so that (x, y)= (0, 1) becomes (x', y')= \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) (because that lies on the line y= x and has distance 1 from the origin) and (x,y)= (1, 0) becomes \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right) (because that lies on y= -x, the line perpedicular to y= x, in the fourth quadrand and has distance 1 from the origin).

    Since a rotation is linear, we must have x= ax'+ by' and y= cx'+ dy'. Taking x= 0, y= 1, x'= \frac{\sqrt{2}}{2}, y'= \frac{\sqrt{2}}{2}, those become (i) 0= \frac{\sqrt{2}}{2}a+ \frac{\sqrt{2}}{2}b and (ii) 1= \frac{\sqrt{2}}{2}c+ \frac{\sqrt{2}}{2}d. Taking x= 1, y= 0, x'= \frac{\sqrt{2}}{2}, y'= -\frac{\sqrt{2}}{2}, those become (iii) 1= \frac{\sqrt{2}}{2}a- \frac{\sqrt{2}}{2}b and (iv) 0= -\frac{\sqrt{2}}{2}c+ \frac{\sqrt{2}}{2}d.

    Adding (i) and (iii) gives 1= \sqrt{2}a so a= \frac{\sqrt{2}}{2}. (i) is the same as b= -a so b= -\frac{\sqrt{2}}{2}.

    Adding (ii) and (iv) gives 1= \sqrt{2}d so d= \frac{\sqrt{2}}{2}. (iv) is the same as d= c so c= \frac{\sqrt{2}}{2}.

    That is, x= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y' and y= \frac{\sqrt{2}}{2}x+ \frac{\sqrt{2}}{2}y.

    Now, the region is bounded by y= x^4 and y= x itself. Obviously, y= x becomes just y'= 0. y= x^4 becomes \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y= (\frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y')^4.

    I will leave it to you to multiply that out. You are now rotating that around the x' axis from x'= 0 to \sqrt{2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rotate Vector
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: February 2nd 2011, 11:11 PM
  2. volumes rotate about y axis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 27th 2010, 06:54 AM
  3. Replies: 6
    Last Post: February 11th 2010, 07:08 PM
  4. how to rotate a curve
    Posted in the Geometry Forum
    Replies: 4
    Last Post: January 27th 2010, 04:32 AM
  5. rotate conic
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 21st 2008, 11:01 AM

Search Tags


/mathhelpforum @mathhelpforum