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Homework Help: 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
  5. Nov 7, 2011 #4


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    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".
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