#### Hckyplayer8

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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?

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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?

What exactly is the problem with the line? Why should it pass through the bounded area?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].

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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.What exactly is the problem with the line? Why should it pass through the bounded area?

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?

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

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

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

R= 2 + x

r = 2 + x

A(y)=(pi)(R

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Are you sure about those radii?So its just the same formula (though integrated in regards to y) with an + 2 displacement?

R= 2 + x^{1/3}

r = 2 + x^{3}

A(y)=(pi)(R^{2})-(pi)(r^{2})

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.

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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?

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Convince me! Explain why you think it

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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.rightof x=0; is the same point x+2 units to theleftof x=2?

Convince me! Explain why you think itwouldbe 2 more.

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I'm with you so far.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?

R = the function that bounds the upper portion of the shaded region = x

r = the function that bounds the lower portion of the shaded region = x

In terms of x, it is delta x.

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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.

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A horizontal distance is always the right endpoint minus the left endpoint, x_{2}-x_{1}. 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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I'm obviously lost.

I'll re-re-re-re-read the section again.

Thank you all for your time.

I'll re-re-re-re-read the section again.

Thank you all for your time.

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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=x^{2} 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

A(y) = (pi)[outer radius)^{2}] - (pi)[(inner radius)^{2}]

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.

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=x

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

A(y) = (pi)[outer radius)

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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Will do.

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

That is the final section we are scheduled to study.

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