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For the region bounded by $y=8-x^3$, the x-axis, and y-axis; determine which of the following is greater: the volume of the solid generated when the region is revolved about the x-axis or about the y-axis.

So I have to find the volume for both regions and I have done this before but with more information, so I am just stuck right now. I am not even sure how to form the definite integral.

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2 Answers 2

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let $V_x, V_y$ be the volumes of the solids generated by rotating the region about the $x, y$ axis respectively. then

$$V_y = 2\pi\int_0^2 x(8-x^3)\, dx = 2\pi(4x^2 - x^5/5)|_0^2 = 59.2\pi$$

$$ V_x = \pi\int_0^2 (8-x^3)^2 \, dx = \pi(64x-4x^4+x^7/7)|_0^2=82.286\pi$$

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At first you compute the intersections of the curve $y=8-x^3$ with the x and y axis by finding the Zeros. You have then the Points $(2,0),(0,8)$. For computing the volume of the revolved solid you can use the equation $V = \pi \int_0^2 y^2 dx$ (Integration interval is Chosen by your intersection points). Moreover you have to compare $V$ with $\int_0^8 x^2 dy$ by solving by $x$.

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