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Hi again John, The region to be revolved runs from $x = 0$ to $x = 4,$ so I don't know why you have the integral from 0 to 2. I get \[V = \pi \int_{0}^{4} [16 x^{3/2}  x^3] dx.\] An antiderivative of $x^{3/2}$ is $\large \frac25 x^{5/2}$ and an antiderivative of $x^3$ is $\large \frac14 x^4$ and hence I get \[\pi \int [16 x^{3/2}  x^3] dx = \pi [16 \times \frac52 x^{5/2}  \frac14 x^4] .\] Evaluating from 0 to 4 gives $V = \large \frac{407 \pi}{3}.$ Penny  


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