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Solutions to problems 1 and 2 from homework #7 in math 1230: calculus 2. The problems involve finding the volume of solids of revolution obtained by revolving the area under the curve y = 1/x^2 around the x-axis and y-axis, and determining the values of p for which the volumes are finite.
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Problem 1 Take the area under the graph of y = (^) x^1 p from x = 1 to x = ∞, and revolve it around the x-axis. For which values of p is the volume of the resulting solid finite?
Solution The volume of the slice at position x is
volume = π
xp
dx
so the total volume is ∫ (^) ∞
1
π
xp
dx = π
1
x^2 p^
dx.
We know from class that this integral converges when 2p > 1 and that it diverges otherwise. We conclude that the volume is finite when p > 12. 2
Problem 2 Take the same region as in Problem 1 and revolve it around the y-axis. For which values of p does that solid have finite volume?
Solution There are two ways to go about this one.
volume = π
y−^ p^1 )^2 − 12
dy
so the total volume is ∫ (^1)
0
π
y−^
(^1) p^ )^2 − 12
dy = π
0
y−^
(^2) p dy − π.
We know from Quiz 4, Problem 2 (in the future!) that this integral converges when − p^2 > −1, and diverges otherwise. Thus the volume is finite for p > 2.
volume = 2πx
xp^
dx
so the total volume is ∫ (^) ∞
1
2 πx
xp^
dx = 2π
1
xp−^1
dx
and we know from class that this integral converges only when p− 1 > 1, which is to say that p > 2.
In either case we see that the volume is finite when p > 2 and infinite otherwise. 2