Using shell method to find the volume of a solid

In summary, the conversation discussed using the shell method to find the volume of a solid generated by revolving a region bounded by a line and a parabola around different lines. The equation v=2pi∫(shell radius)(shell height) was used, and a mistake was made in the calculation of the primitive function at x=-3.
  • #1
mmont012
39
0

Homework Statement



Use the shell method to find the volume of the solid generated by revolving the region bounded by the line y=2x+15 and the parabola y=x2 about the following lines:

a) The line x=5 b) The line x= -3 c) The x-axis d) The line y=25

Note: leave answer in terms of pi

Homework Equations



v=2pi∫(shell radius)(shell height)

The Attempt at a Solution



I know that I am making a mistake somewhere; I have a feeling that it is in my set... I am hoping that someone will be able to point it out to me. I am stuck on part a) and if I figure out my mistake I am confident that I will be able to do the other parts of the problem.

First thing that I did (after graphing which is attached) was find the limits of integration:
2x+15=x2
x2-2x-15
(x-5)(x+3)
So the limits of integration are from -3 to 5.

v=2pi∫(5-x)(2x+15-x2)dx

v=2pi∫10x+75-5x2-2x2-15x+x3

v=2pi∫75-7x2-5x+x3

v=2pi(75x-(7/3)x3-(5/2)x2+(1/4)x4

(plug in the limits of integration)
v=2pi [(2125/12)+(607/4)]

v=2pi(1973/6)

v=1973pi/3 <---this answer is wrong.

I hope that someone will be able to help me with this, thanks for stopping by and sorry for the crappy paint graph.
 

Attachments

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  • #2
mmont012 said:
v=2pi(75x-(7/3)x3-(5/2)x2+(1/4)x4

(plug in the limits of integration)
v=2pi [(2125/12)+(607/4)]

Your setup is fine, but your value for the primitive function at x=-3 is not correct. Try computing that particular value again. :)
 

Related to Using shell method to find the volume of a solid

1. How do you use the shell method to find the volume of a solid?

The shell method is a mathematical technique used to calculate the volume of a solid of revolution. To use this method, you need to integrate the distance from the axis of revolution to the outer edge of the solid, multiplied by the circumference of the shell, over the desired interval. This will give you the volume of the solid.

2. What type of solids can you calculate the volume of using the shell method?

The shell method can be used to find the volume of any solid that can be formed by rotating a curve around an axis. This includes solids such as cylinders, cones, and spheres.

3. What is the difference between the shell method and the disk method?

The shell method and the disk method are both techniques for finding the volume of a solid of revolution. The main difference is that the shell method uses cylindrical shells to approximate the solid, while the disk method uses circular disks. In some cases, one method may be easier to use than the other.

4. How do you set up the integral for the shell method?

To set up the integral for the shell method, you need to identify the axis of revolution and the desired interval of rotation. Then, you need to determine the distance from the axis to the outer edge of the solid, and the circumference of the shell at that distance. Finally, you integrate the product of these two values over the desired interval to find the volume of the solid.

5. Can you use the shell method to find the volume of a solid with a hole in it?

Yes, the shell method can be used to find the volume of a solid with a hole in it. You will need to set up two separate integrals, one for the outer shell and one for the inner shell, and subtract the volume of the inner shell from the volume of the outer shell to find the total volume of the solid.

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