Volume Integration: Calculate Rotated Region Bounded by y=9-X^2, X=2

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Calculate the volume generated when the region bounded by the curve y = 9 - X^2 , the line X = 2 and the x-axis is rotated 2pi about the y-axis.


For the part that lies in the second quadrant, i can integrate 9 to 0, then plug it into the volume formula. Then i have to add the volume of the part in the first quadrant. But since the X = 2 line slices a part of the area off i don't know how to integrate it respective to the y axis.

 

[MathStudent]

Try using the method of cylindrical shells which should be outlined in your book.

Graph the region that is described. This is simple to do,, draw a downward porabola with y intercept at y=9, and a vertical line at x=2. The region you want to consider is the region in the first quadrant that lies under the porabola and over the x-axis and between the lines x=2 and x=3.

hint:
the height of a shell will be 9 - x^2
the radius of the shell is x, so the circumference is 2(pi)x
the thickness of the shell is dx
you need to sum up all the shells with radius from x=2 to x=3

I hope I haven't given too much away.

 

[BobG]

Actually, you want the interval from 2 to 3.

The region would have to also be bounded by x=0 if you were going from 0 to 2.

 

[MathStudent]

Yes, thank you BobG! I overlooked that.

P.S. I will edit my post so as not to cause confusion.