How Does Translating a Region Affect the Volume of a Solid of Revolution?

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SUMMARY

The discussion centers on the relationship between translating a region and the volume of a solid of revolution. Specifically, it confirms that translating the region bounded by y=0, x=9, and y=x/3 up by 2 units, while also translating the axis of rotation from y=-2 to y=0, results in the same volume. The integral used to calculate the volume remains unchanged, affirming that both configurations yield identical results. This principle is crucial for understanding the effects of transformations on geometric properties.

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Homework Statement



Find the area of the region bounded by y=0, x=9, y=x/3 and rotated about y=-2

The Attempt at a Solution



The answer is [tex]\pi[/tex][tex]\int^{9}_{0}[/tex](2+x/3)-2^2dx

I'm just wondering if this is the same thing as saying find the area of the region bounded by y=2 instead of y=0, y=2+x/3, x=9, and rotated about y=0 rather than y=-2. In other words, is everything just moved up by two units?
 
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Yes. If you translate the region up by 2 units, and translate the axis or rotation by the same amount, the two solids of rotation have the same volume. In fact, the integrals turn out exactly the same.
 

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