I’m working through a calculus problem where I need to find the volume of a solid of revolution, and I’ve set up my integral using the washer method. I keep getting a different answer than the textbook, and I think my error is in determining the correct limits of integration for the region bounded by y = x² and y = 2x. How do you properly establish those bounds when the curves intersect?
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How do I set bounds for the washer method with y=x² and y=2x?
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Kyle21
Junior Member
BrianJ
Junior Member
Intersection happens where y equals x squared and y equals two x. That gives x equals zero or x equals two, so the region is between x equals zero and x equals two. If you rotate about the x axis the outer radius is the top value which is two x and the inner radius is the bottom value which is x squared. The volume equals pi times the integral from zero to two of four x squared minus x to the fourth with respect to x. This evaluates to sixty four pi over fifteen.
GregoryCJ
Junior Member
How you set the limits depends on the axis of rotation. If you use washers perpendicular to the x axis you integrate with respect to x from zero to two because those are the intersection x values. If you instead rotate about the y axis you would use y bounds from zero to four and you would have to express x in terms of y such as x equals square root of y and x equals y over two.
Sophia_L
Junior Member
I tried a similar setup and kept getting a different number. On the interval from zero to two the line y equals two x sits above the parabola y equals x squared, so it must be the outer radius. If you swap them you get a smaller result. It happened to me when I mis read which curve is on top.
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