Consider the solid obtained by rotating the region bounded by the given curves about the line y = -1.? y=3/x y?

The cross-section of it is
π(3/x - (-1))^2 - π(-1)^2
= π[(3/x + 1)^2 - 1].

Thus, the volume is
∫ π[(3/x + 1)^2 - 1] dx where x = 1 to 3
= π ∫ (3/x + 1)^2 - 1 dx where x = 1 to 3
= π ∫ 9/(x^2) + 6/x dx where x = 1 to 3
= π(-9/x + 6ln|x| + C) where x = 1 to 3 and C is a constant
= π(-9/3 + 6ln|3| + C) - π(-9/1 + 6ln|1| + C)
= π(-3 + 6ln3) - π(-9)
= 6π + 6πln3
= 6πln(3e) units cubed.

those boundaries set just an empty space that the graph of y = 3/x doesn't occupy. the graph (y = 3/x) only has values in the 1st and 3rd quadrant, whereas your boundaries occupy a space in the 4th quadrant, which is just outside, and looks like a box that you bound.