Volume By Integration

[prettycakemonster]

Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the y-axis:

y=(x-2)^3-2, x=0, y=25

(a)solve by either the disk or washer method
(b)solve by the shell method
(c)state which method is easiest to apply

I don't need someone to solve this for me. I just need help with setting up the equations. I know which equations to use, but I don't know how to set it up with the given information. Please explain to me how I can set up either the disk or washer method for solving this, and the shell method.

 

[HallsofIvy]

Have you graphed this so you know what region is being rotated?
Do you know what the difference between the "disc" and "washer" methods is? (You don't get to choose one or the other.)

A very thin horizontal disc would have a radius extending from the y-axis to y= (x- 2)^3- 2 has what radius and what area?

 

[prettycakemonster]

Have you graphed this so you know what region is being rotated?
Do you know what the difference between the "disc" and "washer" methods is? (You don't get to choose one or the other.)

A very thin horizontal disc would have a radius extending from the y-axis to y= (x- 2)^3- 2 has what radius and what area?

No, I don't really know the difference. My teacher didn't explain it to the class very clearly. What is the difference and how do I know which one I should use?
Also, I don't know how to graph it without a graphing calculator and I won't be able to one the test.

 

[stapel]

I don't know how to graph it without a graphing calculator....

I'm sorry to hear that they never covered graphing quadratics in any of your algebra courses. :shock: Did they teach you any graphing, or do you need to start at the beginning? Thank you!

 

[prettycakemonster]

I'm sorry to hear that they never covered graphing quadratics in any of your algebra courses. :shock: Did they teach you any graphing, or do you need to start at the beginning? Thank you!

Yes, I know basic graphing. I'm just not sure what a good method is that I can use for graphing this.

 

[srmichael]

Yes, I know basic graphing. I'm just not sure what a good method is that I can use for graphing this.

Based on how the function is given to you, (x-2)^3 - 2, it looks like they are assuming that you should know how to transform a parent function, in this case y = x^3, by subtracting 2 from the x value and subtracting 2 from the function itself. The (x-2)^3 part of the function moves the parent function two units to the right and the -2 part of the function moves the parent function down 2. Put it another way, all the points on the y = x^3 function can now be written as (x + 2, y - 2) for the y = (x-2)^3 -2 function.

See if you can sketch this out now to see the region that you will be revolving around the y-axis.

 

[DrPhil]

Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the y-axis:

y=(x-2)^3-2, x=0, y=25

(a)solve by either the disk or washer method
(b)solve by the shell method
(c)state which method is easiest to apply

I need someone to solve this for me. I just need help with setting up the equations. I know which equations to use, but I don't know how to set it up with the given information. Please explain to me how I can set up either the disk or washer method for solving this, and the shell method.

A very thin horizontal disc would have a radius extending from the y-axis to y= (x- 2)^3- 2 has what radius and what area?

Since you are rotating about the y axis, look at disks about that axis. [You only need "washers" if there is a hole in the middle.] The limits on y are from 0 to 25. Setting up requires that you first solve for x as a function of y, since the area of each disk is pi*x^2.

The shell method may be easier to set up. The limits of x are from 0 to the value of x for such that y=25. each cylindrical shell has a radius x, height y(x), and thickness dx.

Please show us how far you can get so we know where you are stuck.

 

[stapel]

To learn how to graph quadratic functions in particular, try here. There are loads of other type-specific graphing lessons, too, like for logs, exponentials, general polynomials, and trig functions. You'll probably want to study up on them at some point soon.

To learn about the "shell" method, try here (it's also called the "cylinder" method). To learn about the "washer" method, try here (it's also called the method of "rings).

You may want to have a serious discussion with your academic advisor, and possibly also your school's administration, about this class that isn't actually teaching the material. :???:

 

[prettycakemonster]

With everyone's help:

I was able to calculate the volume using the shell method. My answer was 1250π. The interval was [0,5].
However, I still can't figure out how to calculate the volume for this using the disk method. I attempted, but I ended up with and irrational number and the answers should be equal to each other. Is my answer using the shell method correct? Also, I still need help with the disk method.

 

[DrPhil]

I was able to calculate the volume using the shell method. My answer was 1250π. The interval was [0,5].
However, I still can't figure out how to calculate the volume for this using the disk method. I attempted, but I ended up with and irrational number and the answers should be equal to each other. Is my answer using the shell method correct? Also, I still need help with the disk method.

Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the y-axis:

y = (x - 2)^3 - 2, x=0, y=25

You have to find x as a function of y.

y + 2 = (x - 2)^3

(y + 2)^{1/3} = x - 2

x = (y + 2)^{1/3} + 2

The area of the disk at height y is πx^2, and if you multiply that area times the incremental thickness dy, then you have incremental volume:

dV = π x^2 * dy = π [(y + 2)^{2/3} + 4(y + 2)^{1/3} + 4] dy

The lower limit for y is the value when x=0, which is -10. The upper limit is given as 25.
If we substitute u = (y+2)^{1/3}, then du = (1/3) (y+2)^{-2/3} dy, or dy = 3 u^2 du.
The limits of integration become y=-10 --> u = -2, and y=25 --> u = 3

\(\displaystyle \displaystyle V = \pi\int_{-2}^{3} \left[u^2 + 4u + 4\right] (3\ u^2)\ du\)

There are lots of intermediate steps for you to check in that substitution! If you agree with what I did, then the actual integration is not too bad. Check my work, and let us know if you agree.

 

[stapel]

And if what you're getting does not agree with what DrPhil has provided, then please include a clear listing of your steps so far. Thank you!