Volumes created by rotations are generally calculated through integration. You first have to determine the 2-D region's boundaries, and then rotate that around a line (the y-axis, in your case). In rotating, you're creating a figure, which you break down by using either disks/washers or cylindrical shells, depending on how complicated the 3-D shape is.
In this case, you need to define the boundaries better. You have y = kx - x^2, which is an inverted parabola with intercepts at (0, 0) and (k, 0) and its vertex at (k/2, (k^2)/4). The region that you're rotating could be the area above the x-axis, or it could be part of that region, or it could be everything to the right of the y-axis. So we need to figure that part out first before we go further.
Once you have your boundaries set, you can set up a problem. It may be easier to use a cylindrical shell setup, since you're going around the y-axis. If you envision your 3-D solid and peel away a shell, the shell is basically a rectangle with some height (depending on the boundaries - if we were to take just the positive part of this parabola, your height would simply be the y value, or kx - x^2), some length (really, a circumference - something like 2*pi*x), and with some thickness, dx. You'd also need some boundaries for the integral (again, assuming that it was the entire positive part of the parabola, your boundaries would be from 0 to k). So your setup would be something like (I'm using S as the integration symbol):
V = S (kx - x^2)(2 * pi * x) dx, from 0 to k
...which is basically a length * width * height.
Hopefully this will get you started. Again, figure out your boundaries and that'll get you going with setting up the integration. Once you've done the integration and plugged in 0 and k, the rest is easy - set it equal to 10 and solve for k. Or, if you wanted to add on to your post to clarify your boundaries, we can help you set everything up.
2007-04-12 03:36:22
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answer #1
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answered by igorotboy 7
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2016-10-28 07:12:13
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answer #2
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answered by ? 4
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