Find the volume of the solid obtained by rotating the region bounded by y=4-x, y=x+2, and y=1

, about the line y=3.

A) 3pi/5
B) 32pi/3
C) 18pi/5
D) 10pi

Thank you.

When y = 1:
1 = 4 - x ==> x = 3, and 1 = x + 2 ==> x = -1.
(So the lines y = 4 - x and y = x + 2 intersect y = 1 at x = 3 and x = -1, respectively.)

Next, we check that y = 4 - x and y = x + 2 intersect when
4 - x = x + 2 ==> x = 1, which is third and final vertex of the region.

Plot:
http://www.wolframalpha.com/input/?i=plo…

It is easier to think of this region as that between x = y - 2 and x = 4 - y with y in [1, 3].
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So, taking a horizontal strip in the region at some y in [1, 3] and rotating it about y = 3 yields a cylindrical shell with radius 3 - y and height (4 - y) - (y - 2) = 6 - 2y.

Hence, the volume equals
∫(y = 1 to 3) 2π (3 - y)(6 - 2y) dy
= ∫(y = 1 to 3) 4π (3 - y)(3 - y) dy
= ∫(y = 1 to 3) 4π (y - 3)^2 dy
= (4π/3)(y - 3)^3 {for y = 1 to 3}
= 32π/3.

I hope this helps!