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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.

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Applications of Integration

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mhm. Okay. This integral represents a volume and because of the way it's written, you can tell that it's big. R squared minus little r squared mm So this is a washer. All right. So um and since it's dy we're going around the y axis and we got a washer. Okay, so big are is from here to here. So big are is big. R squared is y to the fourth. So big are is y squared. Okay. But really it's X on the right is what I'm going to call it X on the right, which is why squared? So why is the square root of X? Okay. And then little R is the distance from here to here And we have little r squared equals coach. Should be white to the 8th Equals Way to the 8th. So little are is white the 4th and that's X. On the left. So to draw that we draw wife was The 4th Root of X. Okay. And then why is going from 0 to 1? Okay, so This is 0-1. So um We know it goes through 00 and 11. They both do. Okay, so I picked another point that I knew both the square root and the fourth root of 16. What? 1/16? So the square root of 1 16th? It's 1 4th But the 4th root of 1/16 is 1/2. Okay, so that puts The 4th root up on the top because a half is more than 1/4. So this is the fourth root of X. And then here's the square root of X. And then we cut slices in it this way and sent them around. Okay? So yeah, its shape is like a little funnel looking thing, okay, with the outer edge being like was the square root of X. And the whole in there being like was the 4th root of X and it's 0 to 1 and it's rotated around the Y axis.

Oklahoma State University

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Applications of Integration

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