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# Shell Method [Solved!]

### My question

Under the section, "When to use the shell method", with the first example, should the disks (blue rectangles) be horizontal rectangles located between the graph and line x = 1?

### Relevant page

Shell Method | Brilliant Math & Science Wiki

### What I've done so far

Not clear as to how they determined the region to subtract is the region bounded by y = 1,
y = sqrt x, and the y-axis.

X

URL:<a href="https://brilliant.org/wiki/shell-method/">Shell Method | Brilliant Math &amp; Science Wiki</a>

Under the section, "When to use the shell method", with the first example, should the disks (blue rectangles) be horizontal rectangles located between the graph and line x = 1?
Relevant page

<a href="https://brilliant.org/wiki/shell-method/">Shell Method | Brilliant Math &amp; Science Wiki</a>

What I've done so far

Not clear as to how they determined the region to subtract is the region bounded by y = 1,
y = sqrt x, and the y-axis.

## Re: Shell Method

What they've done in their example is to find the volume of the cylinder generated when rotating the unit square (with vertices [0,0], [1,0], [1,1] and [0,1]) around the y-axis, then they've subtracted the volume of the cone-shaped part from that, to give the required funnel-shaped volume.

So their blue rectangles are correct how they have them (although I'd normally only have one rectangle, as a "typical" rectangle.)

We could have found that volume if the blue rectangle(s) is(are) where you thought it(they) should be (forming the funnel-shaped volume).

In that case, we have to proceed as follows:

For the function we need to integrate, we take the outer border (which is x=1, the line parallel to the y-axis) minus the inner border (the curve y=sqrt(x) that is, x=y^2)

Volume = pi int_0^1 (1-(x)^2)dy

= pi int_0^1 (1-y^4)dy

= pi [y-y^5/5]_0^1

= pi[(1-1/5)-(0-0)]

= (4pi)/5

It amounts to the same thing as their answer, just that it's written in one expression.

X

What they've done in their example is to find the volume of the cylinder generated when rotating the unit square (with vertices [0,0], [1,0], [1,1] and [0,1]) around the y-axis, then they've subtracted the volume of the cone-shaped part from that, to give the required funnel-shaped volume.

So their blue rectangles are correct how they have them (although I'd normally only have one rectangle, as a "typical" rectangle.)

We could have found that volume if the blue rectangle(s) is(are) where you thought it(they) should be (forming the funnel-shaped volume).

In that case, we have to proceed as follows:

For the function we need to integrate, we take the outer border (which is x=1, the line parallel to the y-axis) minus the inner border (the curve y=sqrt(x) that is, x=y^2)

Volume = pi int_0^1 (1-(x)^2)dy

= pi int_0^1 (1-y^4)dy

= pi [y-y^5/5]_0^1

= pi[(1-1/5)-(0-0)]

= (4pi)/5

It amounts to the same thing as their answer, just that it's written in one expression.