## 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.

$y=\sqrt{1-x^3}, \quad y=0, \quad x=0$

Solutions

Verified**b.**

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

$\begin{aligned} V&=2\pi \int_a^b p\left(x\right)h\left(x\right)dx && \text{Apply Shell method.} \\ &=2\pi \int_0^1 x\sqrt{1-x^3} dx && \text{Substitute} \ x \ \text{for} \ p\left(x\right) \ \text{and} \ \sqrt{1-x^3} \ \text{for} \ h\left(x\right). \\ &\approx 2\pi \cdot 0.37 && \text{Use the IntegralCalculator.} \\ &\approx 2.32 \end{aligned}$

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

$y=\sqrt{1-x^3},\text{ }x=0,\text{ }y=0,$

around the $y$-axis.

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

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