How do you find the volume of the solid generated by revolving the region bounded by the graphs of the equations #y=sec(x)# , y=0, #0 <= x <= pi/3# about the line y = 5?
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To find the volume of the solid generated by revolving the region bounded by the graphs of the equations y=sec(x), y=0, 0 <= x <= pi/3 about the line y=5, you can use the disk method.

First, find the points of intersection of the two curves. Set sec(x) equal to 0 and solve for x to find the xcoordinate of the intersection point.

Next, integrate the area of the crosssections perpendicular to the axis of revolution. Since the region is bounded by the xaxis and y=sec(x), the radius of each disk is 5  sec(x).

The integral setup is ∫[0,π/3] π(5  sec(x))^2 dx.

Evaluate the integral to find the volume.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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