1. Limited time only! Sign up for a free 30min personal 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!

Integration (Enclosed Area & Rotation Volume)

  1. Nov 3, 2011 #1
    1) Find the area enclosed by the curve x=3+cos(t) and y=4sin(t).

    My tutor gave me a hint which confuses me.

    A=-4 integral y dx where the upper limit is pi/2 and the lower limit is 0.

    I do not understand the -4 part. why must we multiply the integral by -4?



    2) Find the volume of the solid generated by revolving about the y-axis the area between the first arc of the cycloid x=t-sin(t), y=1-cos(t) and the x-axis. Use the shell method.

    v=2pi integral xy dx
    =2pi integral (t-sin(t)) (1-cos(t))^2 dt

    I'm stuck at this point.
     
  2. jcsd
  3. Nov 4, 2011 #2
    could anyone help me with these two questions?
     
  4. Nov 7, 2011 #3
    anyone?=/
     
  5. Nov 7, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If x= 3+cos(t) and y= 4 sin(t), then cos(t)= x- 3 and sin(t)= y/4 so that [itex](x- 3)^2+ y^2/16= sin^2(t)+ cos^2(t)= 1[/itex]. The figure is an ellipse with center at (3, 0), minor semi-axis on the x-axis, of length 1, and major semi-axis parallel to the y-axis, of length 4. The area of such an ellipse is [itex]\pi[/itex] times the product of the two semi-axes, here, [itex]4\pi[/itex].

    A "standard" parameterization for a circle would be "x= R cos(t), y= R sin(t)" where R is the radius and t goes around the full circle, from 0 to [itex]2\pi[/itex]. Using different "R" for x and y (1 and 4) gives an ellipse rather than a circle. Going from 0 to [itex]\pi/2[/itex] gives you 1/4 of the ellipse. Because of the symmetry, multiplying that by 4 will give the area of the entire ellipse. The negative is because with x= R cos(t) (or x= cos(t)+ 3), [itex]dx= -R sin(t)dt[/itex].

    You can also see this as a special case of Greens Theorem:
    [tex]\int\int_D \left(\frac{\partial M}{\partial x}- \frac{\partial N}{\partial y}\right) dxdy= \oint Ndx+ Mdy[/tex]

    In particular, if we take M= 0 N= y the left side becomes [itex]\int\int_D -1 dxdy[/itex] which is just the negative of the area of region D. The right side becomes [itex]\oint y dx[/itex] where the integral is around the circumference of the region.

    However, this problem has nothing to do with "volume of rotation" or even "volume".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration (Enclosed Area & Rotation Volume)
  1. Volume enclosed (Replies: 1)

Loading...