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Homework Help: Volume of revolution

  1. Mar 17, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm having a bit of trouble when it comes to volume of revolutions and areas. I find it quite difficult when it comes to setting up the integral. I'm not sure when to use the shell or washer method. Could someone explain to me or give me a tutorial on how to set up the equations thanks!

    Here are a few examples

    1)Find the volume of the solid obtained by rotating the region enclosed by the curves y=x^2 , x = 3, x = 8, and y=0 about the line x=9.

    2)The region enclosed by the curves y = x^2 and x = y^2 is rotated about the line y = -2. Find the volume of the resulting solid.

    3)Find the volume of the solid formed by rotating the region enclosed by the curves y=e^(x) + 2, y=0 , x=0, and x=0.1 about the x-axis.

    4)Find the volume of the solid obtained by rotating the region enclosed by the curves y=x^2 and x = y^2 about the line x=-1.

    5)The region enclosed by the curves x = 1 - y^4 and x = 0 is rotated about the line x = 4. Find the volume of the resulting solid.

    Thanks for all the help!

    2. Relevant equations

    regular integration equations?

    3. The attempt at a solution

    1) (integral 9 top 3 bottom) pi(81-x^4)dx

    2) (integral 1 top 0 bottom) 2pi(x^2 - x^.5)(2+x)dx

    3) (integral 1 top 0 bottom) pi((e^(x) + 2)^2)dx

    4) (integral 1 top 0 bottom) 2pi(x^2 - x^.5)(1+x)dx

    5) not sure

    thanks for all the help!
     
  2. jcsd
  3. Mar 18, 2010 #2
    here's a few pointers that might help you out with solids of revolution.

    -for washer method type problems, you're always integrating with respect to the axis opposite the one you're rotating; example, say youre taking a certain area and rotating it around the line x=1. it is a vertical line, like the y axis, so in that case you are integrating with respect to x. likewise, if you're rotating around something like y=8/5, you're rotating around an axis much like the x axis, so you would be integrating with respect to y.

    -for shells, its just the opposite. if youre rotating around a vertical line, like x=1, youre integrating with respect to y.

    -another thing you'll have to do is draw a picture. for many of these types of problems, the instructions dont say which method you should use. many times its a matter of ruling out based on how the cylinders or shells are being stacked. then sometimes you get a problem where you could use both methods; if such a thing happens, and there are no specific directions, use whichever one youre most comfortable with. usually something along the lines of radii being bounded by the same function will rule out either one of the methods or both.

    -for washers, the smaller radius, r, will be from your axis of rotation to the closer of the two functions; likewise, the big radius, R, is the function that is farther away.

    -for shells, the height is the distance between the two functions given; sometimes it could just be one function and then a certain line, or maybe an axis. the Radius is different in that it is just based on the distance from the axis of rotation to the outermost shell.

    -sometimes the issue is just trying to visualize the equations. ones that can be tough are functions that are y polynomials, in which you can't just solve for y to plug into a calculator. it might be a good idea to download a program that can graph functions of y.
    not a super useful tip, but worth mentioning.

    overall, get into the habit of drawing a very detailed graph so you can see what's going on. that itself is the best advice for doing these problems.

    hope this helped!
     
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