# Applications of Integration: Volumes Part II

#### Hckyplayer8

##### Full Member
Sketch the region in the first quadrant enclosed by y =x^3 and y = x^1/3 and find the volume of the solid produced when that region is revolved about the line x = 2.

I sketched and shaded the region. This question doesn't make much sense to me because x=2 is outside the bounded area of [0,1].

Would I just add 1 to my functions?

#### lev888

##### Senior Member
I sketched and shaded the region. This question doesn't make much sense to me because x=2 is outside the bounded area of [0,1].
What exactly is the problem with the line? Why should it pass through the bounded area?

#### Hckyplayer8

##### Full Member
What exactly is the problem with the line? Why should it pass through the bounded area?
I must be thinking wrong. All the examples I have done up to this point have had the region touching the axis that it is being revolved around.

So when I shaded this region in, it threw me off.

So this problem is just integrating in respect to y of the vertical line created when x = 2?

#### Dr.Peterson

##### Elite Member
You'll be using either washers or cylindrical shells whose centers are on x=2. The radii you use will be distances from that line.

I would expect that you have seen examples of revolution about a line other than the axis.

#### Hckyplayer8

##### Full Member
You'll be using either washers or cylindrical shells whose centers are on x=2. The radii you use will be distances from that line.

I would expect that you have seen examples of revolution about a line other than the axis.
So its just the same formula (though integrated in regards to y) with an + 2 displacement?

R= 2 + x1/3
r = 2 + x3

A(y)=(pi)(R2)-(pi)(r2)

#### Jomo

##### Elite Member
For now on can you please upload a picture with the way you labeled everything so that you can get perfect help? Thanks.

• Hckyplayer8

#### Dr.Peterson

##### Elite Member
So its just the same formula (though integrated in regards to y) with an + 2 displacement?

R= 2 + x1/3
r = 2 + x3

A(y)=(pi)(R2)-(pi)(r2)

Check against the drawing you're going to show us, and specifically compare them with the radii to a particular point, maybe (0,0) or (1,1). Also, explain to us why you wrote those expressions, and how they relate to the drawing.

• Hckyplayer8

#### Hckyplayer8

##### Full Member
I'll upload my pic but I already double checked it on Mathway.

If the previous radii were simply the functions rotated around x = 0 and this problem requires the central axis to be x=2, wouldn't the radii just be the functions +2 as I proposed?

#### Dr.Peterson

##### Elite Member
No. A point on either curve is x units to the right of x=0; is the same point x+2 units to the left of x=2?

Convince me! Explain why you think it would be 2 more.

#### Hckyplayer8

##### Full Member
No. A point on either curve is x units to the right of x=0; is the same point x+2 units to the left of x=2?

Convince me! Explain why you think it would be 2 more.
Oh. So if x=2 acts as a proxy of x=0 then point [1,1] is actually -1 and point [0,0] is -2.

#### Hckyplayer8

##### Full Member
Unfortunately, you didn't draw in the important bit: What ARE r and R?

I've drawn in an x-coordinate and R for one point:
View attachment 15175
The blue segment is R, right? (the distance from the axis to a point on the outer curve) What is it, in terms of x?
I'm with you so far.

R = the function that bounds the upper portion of the shaded region = x1/3
r = the function that bounds the lower portion of the shaded region = x3

In terms of x, it is delta x.

#### Dr.Peterson

##### Elite Member
No! Almost nothing you say here is correct.

In your integrand, R and r are radii -- distances from the axis x=2 to a point on the curve. They are not the functions themselves, and they are not "upper" and "lower". They are "left" and "right". They have nothing to do with delta x; the formula you are using is for the washer method, where you will be integrating with respect to y, not x, so there is no delta x in sight anywhere!

Please try sketching the typical "washer" from which you will obtain the volume. Textbooks I know typically tell you to do that before you do anything else. If you haven't done so yet, find an example in your textbook or elsewhere and follow it.

• Hckyplayer8

#### Jomo

##### Elite Member
A horizontal distance is always the right endpoint minus the left endpoint, x2-x1. In your case you know that the right endpoint is 1. You need to find the two (one for each radius) left endpoints. Do you want them in terms of y or in terms of x?

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#### Hckyplayer8

##### Full Member
I'm obviously lost.

Thank you all for your time.

• Jomo

#### Hckyplayer8

##### Full Member
I believe the closest example in my book (Stewart's Single Variable Calculus 7th edition) is as follows-

Page 358, Example 6

"Find the volume of the solid obtained by rotating the region in Example 4 about the line x = -1."

Example 4's problem was

"The region R enclosed by the curves y=x and y=x2 is rotated about the x-axis. Find the volume of the resulting solid"

Example 6's Solution

"Figure 11 shows a horizontal cross-section. It is a washer with inner radius 1+y and outer radius 1+ sqrt(y), so the cross-sectional area is

It then goes on to show the math and figure.

The figure has the line x = -1 drawn which I have replicated by drawing the line in my problem of x=2.

To account for the central axis shift from 0 to -1, the example problem adds 1 to the radii.

I followed that (first incorrectly by adding 2) by subtracting 2. This part I feel pretty confident I have right. Which means my error has to be related to the radii.

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#### Jomo

##### Elite Member
Here and Here

• Hckyplayer8

#### Hckyplayer8

##### Full Member
Here and Here
Will do.

Are you implying this question is in regard to the shell method?

That is the final section we are scheduled to study.

#### lev888

##### Senior Member
To account for the central axis shift from 0 to -1, the example problem adds 1 to the radii.
Do you need to account for the fact that the axis of revolution is now a line parallel to the y axis instead of x?

• Hckyplayer8