Volume of Revolution: y=5, y=x+4/x; x=-1

In summary, the problem involves finding the volume of a solid formed by rotating the region bounded by y=5 and y=x+4/x around the axis x=-1. The student attempted to solve it using the shell method, but made a mistake in the calculation and got the incorrect answer of 8pi(3-ln4). After realizing the error, the student corrected it and found the correct answer.
  • #1
mace2
101
0

Homework Statement


"The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method."

y=5, y=x+4/x; about x=-1


Homework Equations


Upon plotting it I decided it would be best to use the shell method. I'm not sure how to express it in your fancy symbols but I included all of my work as an attachment.


The Attempt at a Solution


Please see attachment. The correct answer is actually 8pi(3-ln4). I have gone over it a few times and I don't see what I did wrong. Any advice whatsoever would be great, thanks.

P.S. Sorry the image is grainy, I didn't know how to make it small enough to fit the guidelines and still be legible.
 

Attachments

  • Math 001.GIF
    Math 001.GIF
    230.7 KB · Views: 480
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  • #2
Ah I just figured it out. I accidentally wrote 5-x+4/x, when i really meant 5-(x+4/x).

Stupid parantheses!
 
  • #3
Perhaps it's not the parentheses that are stupid!:rolleyes:
 

1. What is the volume of revolution for the given equation?

The volume of revolution can be calculated using the formula V = π∫(upper limit) (lower limit) (radius)^2 dx, where the radius is the distance from the axis of revolution to the curve at each point. In this case, the upper limit is x = 5 and the lower limit is x = -1.

2. How do you find the radius for this equation?

The radius can be found by taking the difference between the two given equations, which in this case is y = 5 and y = x + 4/x. Therefore, the radius is r = 5 - (x + 4/x) = (5x - x^2 - 4)/x.

3. Is it necessary to use the upper and lower limits when calculating the volume of revolution?

Yes, the upper and lower limits are essential in determining the volume of revolution as they specify the range of values over which the integral is evaluated.

4. Can the volume of revolution be negative?

No, the volume of revolution cannot be negative as it represents the space enclosed by the curve and the axis of revolution, which cannot have negative values.

5. How does changing the upper and lower limits affect the volume of revolution?

Changing the upper and lower limits will result in a different volume of revolution as it alters the range over which the integral is evaluated. Generally, a larger range will result in a larger volume of revolution, while a smaller range will result in a smaller volume.

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