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# Each integral represents the volume of a solid. Describe the solid.$\pi \displaystyle \int_{0}^1 (y^4 - y^8) dy$

## $\pi \int_{0}^{1}\left(y^{4}-y^{8}\right) d y=\pi \int_{0}^{1}\left[\left(y^{2}\right)^{2}-\left(y^{4}\right)^{2}\right] d y$ describes the volume of the solid obtained by rotating the region$\mathcal{B}=\left\{(x, y) | 0 \leq y \leq 1, y^{4} \leq x \leq y^{2}\right\}$ of the $x y$ -plane about the $y$ -axis.

#### Topics

Applications of Integration

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### Video Transcript

we know the formula were usually looking at is pi times the radius squared, right? So in this context we have pi times the outer radius squared mice the inter radius squared Which is why we have to separate terms in this integral over here. Therefore, what we know is that it is the region between the two curves. So we mentioned there were two. Why did the fourth and why squared between zero and one for why? Therefore, given that we have why we know this region rotates around the y axis.

#### Topics

Applications of Integration

Lectures

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