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

[armolinasf]

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 $$\pi$$$$\int^{9}_{0}$$(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?

 

[Mark44]

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.