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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?

Answer 1

Scarlett Allison

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Answer 2

Samantha Adkison

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 x-coordinate of the intersection point.
Next, integrate the area of the cross-sections perpendicular to the axis of revolution. Since the region is bounded by the x-axis 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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Answer from HIX Tutor

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.

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 &lt;= x &lt;= pi/3# about the line y = 5?

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?