Find the volume of the resulting solid if the region under the curve from to is rotated about the x-axis. Would the volume of this just be the integral from 0 to 1?
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Originally Posted by cdlegendary Find the volume of the resulting solid if the region under the curve from to is rotated about the x-axis. Would the volume of this just be the integral from 0 to 1? no. the integral of the function gives you the net area under the curve. use the disk method. $\displaystyle V = \pi \int_0^1 \left( \frac 1{x^2 + 3x + 2} \right)^2~dx$
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