# xy = 1, x = 0, y = 1, y = 3 Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the y=-2. To determine the bounded region, graph the given equations.(See Figure 1). Then, shade the bounded region.

Since the axis of rotation is horizontal, in cylindrical method, the rectangular strip to be drawn inside the bounded region should be parallel to the axis of rotation. So, draw a horizontal strip inside...

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To determine the bounded region, graph the given equations.(See Figure 1). Then, shade the bounded region.

Since the axis of rotation is horizontal, in cylindrical method, the rectangular strip to be drawn inside the bounded region should be parallel to the axis of rotation. So, draw a horizontal strip inside the bounded region.

When the strip is rotated about y=-2, a cylinder is formed. (See Figure 2). To get its volume using the method of cylindrical shell, the formula is:

`V = 2 pi r h Delta r `

where r is the radius, h is the height and `Deltar` is the thickness of the strip.

So, when the axis of rotation is horizontal, this formula becomes:

`V = int_a^b (2 pi r h) dy`

Base on the figure, the expression that represents the radius of the cylinder is:

`r = y - (-2) = y + 2`

And the height of the strip is:

`h = x - 0 `` `

`h= x`

Since the integral must be expressed in y variable, the x in the equation xy=1 must be isolated. This becomes x=1/y. Plugging this to the height of the cylinder, the h becomes:

`h=1/y`

So, the integral needed to compute the volume of the solid formed when the bounded region is rotated about y=-2 is:

`V= int_1^3 (2 pi * (y+2) * 1/y)dy`

This, integral simplifies to:

`V= 2 pi int_1^3 (1+2/y) dy`

Evaluating this, the result is:

`V = 2 pi (y + 2ln|y|)_1^3`

`V = 2pi [ (3 +2 ln|3|) - (1+2ln|1|)]`

`V = 2pi(2 + 2ln|3|)`

`V = 26.37`

Therefore, the volume of the solid formed when the bounded region is rotated about y=-2 is 26.37 cubic units.

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