Volume of Solid of Revolution: Trash Can Problem

In summary, to find the volume of a silo-shaped trash can, you can use the solid of revolution method by first representing the trash can as a vertical line and a quarter circle. Then, you can integrate the volume of the disks formed by rotating the curve around the y-axis. It is best to split the integration into two parts, one for the constant vertical line and one for the quarter circle. The equations for the curves would be x=15 and x=√(225-(y-76)^2), with y going from 0 to 76 and from 76 to 91, respectively.
  • #1
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Homework Statement


I have to go around and find the volume of a silo-shaped trash can using solid of revolution
height 91cm
Circumference 119.3cm
Diameter 15cm
http://common.csnstores.com/United-Receptacle-European-Designer-15-Gal.-Round-Top-Receptacle~img~UR~UR1180_l.jpg is what the trash can looks like.

I need list of the equations(s) of the curve(s) which are going to be revolved around an axis. I also need to state the upper and lower limits for the equation(s). Then intergrate the curve(s)

Homework Equations


I want to use [tex]\int[/tex] pi r^2 dx and from 0 to r
then do the volume of a sphere and divide by half.
But I believe this is wrong.


The Attempt at a Solution


I tried that equation I wrote and was not right. I think I have to use the shells intergration but don't know the set up.
 
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  • #2
It's hard to understand what you are doing. What I make of this is that your trash can would be represented, on cordinate system, as a vertical straight line, at x= 15, from (15, 0) to (15, 91-15)= (15, 76) and then a quarter circle with center at (0, 76), radius 15, so that its ends are (15, 76) and (0, 91). Rotated around the y-axis, the solid generated is a cylinder topped by a hemisphere.

But you start talking about doing this as a solid of revolution but then say "then do the volume of a sphere and divide by half." Certainly the easiest way to do this would be to use the formulas for volume of a cylinder and sphere, but that is not doing it "as a solid of revolution". I would not use "shells" (and integrating with respect to x), "disks" (and integrating with respect to y) is much simpler. At each y, the radius of a "disk" is the x value at on the curve and the area is [tex]\pi x^2[/tex] so the volume is [tex]\pi r^2 dy[/itex]. That's what you want to integrate as y goes from the bottom to the top of the figure. Because the formula changes at y= 76, it is probably best to do this at as two separate integrals. For the first x= 15, a constant, while y goes from 0 to 76. For the second, the equation of a circle with center at (0, 76), radius 15, is [tex]x^2+ (y- 76)^2= 225[/tex] so [tex]x= \sqrt{225- (y- 76)^2}[/itex] and y goes from 76 to 91.
 

Related to Volume of Solid of Revolution: Trash Can Problem

What is the "Trash Can Problem" in regards to volume of solid of revolution?

The "Trash Can Problem" is a common math problem that involves finding the volume of a three-dimensional shape created by rotating a two-dimensional shape around an axis. In this specific scenario, the two-dimensional shape is a rectangle representing the cross-section of a trash can, and the axis is the height of the trash can.

How do you approach solving the "Trash Can Problem"?

To solve the "Trash Can Problem", you first need to find the equation of the curve created by the rotating rectangle. This can be done by using the formula for circumference of a circle and substituting the height of the rectangle as the radius. Then, you can integrate the equation using the limits of the height of the trash can to find the volume of the solid of revolution.

What is the significance of the "Trash Can Problem" in real life?

The "Trash Can Problem" is a practical application of the concept of volume of solid of revolution. It can be used to calculate the volume of various objects such as bottles, cans, or other cylindrical containers, which is useful in industries such as packaging and manufacturing.

What are some common mistakes when solving the "Trash Can Problem"?

One common mistake when solving the "Trash Can Problem" is forgetting to use the correct units when finding the equation for the curve. Another mistake is not setting up the integral correctly, which can result in an incorrect final answer. It is important to double check the setup and calculations to avoid these errors.

Are there any alternate methods for solving the "Trash Can Problem"?

Yes, there are alternate methods for solving the "Trash Can Problem" such as using the disk or washer method. These methods involve slicing the solid of revolution into thin disks or washers and using the formula for volume of a cylinder to find the volume. However, the shell method is often preferred as it requires only one integral instead of two.

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