Question about Texas Instruments TI-89 Calculator

# How do I find the volume of the solid of revolution rotated around the line y=4 of the area between y=x and y=x^2-2x ?

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Posted on Apr 16, 2014

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Assuming the region of interest is from x=0 to x=3 (unless something is specified, there is no solution), the solution on the TI-89 is found by placing the following on the entry line and pressing ENTER:
PI*S((4-x^2+2*x)^2-(4-x)^2,x,0,3)
where S=integral operator
I get this formula by dividing the volume of interest into an "infinite" number of annular "disks" getting the areas of each "disk" as the difference between the parabolic curve and the straight line curve using the usual area formula for a circle. Summing up all "disk" areas times dx, the thickness of each disk leads to the above integral. The answer should be 98.9601685881 cubic units, barring any goofs.

Posted on Apr 16, 2014

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1. The intersection points of the line y=x, and the parabola y=x^2-2x are (0,0) and (3,3).
2. Calculate the volume generated by the rotation of the arc of parabola around the line y=4, between the intersection points (0,0) and (3,3).
3. Calculate the volume of the truncated cone generated by the rotation of the segment of the line y=x about the line y=4. Use plane geometry mensuration formulas.
4. Subtract the volume of the cone from the volume generated by the arc of parabola.

Posted on Apr 16, 2014

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