# How do you find the volume of the region bounded by #y=6x# #y=x# and #y=18# is revolved about the y axis?

Use washers. to get

The region is the bounded region in:

graph{(y-6x)(y-x)(y-0.0001x-18) sqrt(81-(x-9)^2)sqrt(85-(y-9)^2)/sqrt(81-(x-9)^2)sqrt(85-(y-9)^2) = 0 [-28.96, 44.06, -7.7, 28.83]}

(Steps omitted because once it is set up, I think this is a straightforward integration.)

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To find the volume of the region bounded by the curves (y = 6x), (y = x), and (y = 18) revolved about the y-axis, you can use the method of cylindrical shells.

The limits of integration will be determined by the points where the curves intersect. Setting (6x = x) to find the intersection point:

[6x = x]

[5x = 0]

So, (x = 0) is the intersection point.

The radius (r) of the cylindrical shell at (x) is (x), because the axis of rotation is the y-axis. The height (h) of the cylindrical shell at (x) is the difference between the (y)-values of the upper and lower curves, which is (18 - 6x) - (x = 18 - 7x).

The volume (V) of each cylindrical shell is given by (V = 2\pi rh).

Therefore, the total volume is given by the integral:

[V = \int_{0}^{6} 2\pi x(18 - 7x) , dx]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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