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Find the volume of the solid formed when the region enclosed by the curve 16/x - x^2 , the x axis, and the lines x = 1 and x = 2 is rotated 360 deg about the x axis.

I found the volume of the rectangle that is enclosed in the region rotated about the x axis using the formula pi(r^2)h with r = 4 and h = 1. Then using integration I found the volume of the leftover region rotated between the bounds x =1 and x =2 using integration, summing up both the volumes found gives me 511pi/5. However, the correct answer does not take into the account the volume of the rectangle rotated. Am I struggling to visualise the problem?

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You can compute that volume by the disk method:$$\int_1^2\pi f^2(x)\,\mathrm dx=\frac{431}5\pi.$$

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  • $\begingroup$ Yes I used this method but when I graph the curve, the line x =1 and the line x =2 I get a region where there is a rectangle so I found the volume of the rectangle rotated and then added it to 431pi/5 $\endgroup$ Apr 20 at 10:53

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