Find the volume bound by the curves y=4x-x^2 and y=3, rotated about x=1.

Question

anonymous · Answer

I used the cylindrical shell method and took my circumference to be $$2\pi(1+x)$$ and my height to be $$4x-x^2-3$$ and then I integrated the the product of the two equations from [1,3]. I come up with 8pi, but the answer is 8pi/3. I'm not sure what I'm doing wrong...

anonymous · Answer

Just wondering; are you not told that one of the coordinate axis is bounding your region?

anonymous · Answer

If not, then the region to be considered is the one you mentioned in your post.  Also, if you draw out this region, you should see that the radius of each shell should be (x-1), not 1+x since x is being measured from the y-axis.  Thus, the volume integral (by cylindrical shells) should be $\large\displaystyle 2\pi\int_1^3(x-1)(4x-x^2-3)\,dx$.

Does this clarify things? :-)

anonymous · Answer

Ah! I was so close! I see now why it should be (x-1). Thanks for the help! :)

anonymous · Answer

Oh, and to answer your question, it did not say that any of the axises were bounding the curve.