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# Suppose that a region $\Re$ has area $A$ and lies above the x-axis. When $\Re$ is rotated about the x-axis, it sweeps out a solid with volume $V_1$. When $\Re$ is rotated about the line $y = -k$ (where $k$ is a positive number), it sweeps out a solid with volume $V_2$. Express $V_2$ in terms of $V_1$, $k$, and $A$.

## $V_{2}=V_{1}+2 k \pi A$

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Applications of Integration

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We're told that a region R has an area A. And lies above the X axis. We're told when the region R is rotated about the X axis, It sweeps out a solid with the volume v. one. When the region R is rotated about the line Y equals negative K. For K is a positive number. So a horizontal line below the X axis. It sweeps out of solid with the volume V two were asked to express the second volume V two in terms of the first volume V. One. That's what. Okay. And the area a the region are my digs. Shit, can't get home. It's only three long. Well, first we'll let F of X be the upper border now. So in G Fx be the lower border of our region are yeah, 35 years old. Therefore F of X is greater than G. Fx. Yes. On the interval we'll just call it a B. Okay? Yeah. Yeah. Mhm. Now we know that our region are has an area A. This means that A. Is the integral from A to B of F of x minus G fx D F X, you said. And the region R is rotated about the X axis has a volume. The one using the disk method. So by the disk method we have that The one is equal to. Yes, pi times the integral from A to B of the outer radius F of X squared minus the inner radius G of X squared dx. So I just now if the region R is rotated about the line Y equals negative K. Once again using the washer method, my former they paid some how many, how many we have that are second volume V two is pi times integral from A to B of the outer radius. This is ffx minus negative K squared minus the inner radius, which is G of x minus negative K squared D X. This simplifies to high times. The integral from A to B of this is F of X, which is squared minus G F x squared plus two KF of X -2 Kg FX. Talk honestly put them on. That's We can split this up into two integral. This is the same as pi times and a girl from A to B of F of X squared minus G of X squared dx plus pi times two K. I guess maybe you could do to Cape I. Yeah. Times. And a girl from A to B of ffx minus G of X. Dx. We recognize that the first expression is the same as our Volume V one. And the second expression is the same as two K Pi Times Our Area of the Region R. A. Yeah. The west side. Yeah. Yeah. Walk. Yes. Right up there. Therefore we have that are volume V two is equal to volume V one plus two K pi A. Oh yes, yeah. In this way we can find the volume of the second solid without even using calculus.

Ohio State University

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Applications of Integration

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