## Related questions with answers

Use the integration capabilities of the graphing utility to approximate the volume of the solid generated by revolving the region about the y-axis.

$y=\sqrt[3]{(x-2)^2(x-6)^2}, \quad y=0, \quad x=2, \quad x=6$

Solutions

Verified**b.**

$V=2\pi \int_a^b p\left(x\right)h\left(x\right)dx$

$\begin{aligned} V&=2\pi \int_a^b p\left(x\right)h\left(x\right)dx && \text{Apply Shell method.} \\ &=2\pi \int_2^6 x\sqrt[3]{\left(x-2\right)^2\left(x-6\right)^2} dx && \text{Substitute} \ x \ \text{for} \ p\left(x\right) \ \text{and} \\ &\text{} && \sqrt[3]{\left(x-2\right)^2\left(x-6\right)^2} \ \text{for} \ h\left(x\right). \\ &\approx 2\pi \cdot 29.82 && \text{Use the IntegralCalculator.} \\ &\approx 187.25 \end{aligned}$

In this task, we need to use the recommended software to find the volume of the solid. We are given that the solid is generated by the region bounded by

$y=\sqrt[3]{\qty(x-2)^2\qty(x-6)^2},\text{ }y=0,\text{ }x=2,\text{ }x=6$

around the $y$-axis.

*How can we calculate the volume of the solid of revolution?*

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