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Shell Method [Solved!]

My question

URL:Shell Method | Brilliant Math & Science Wiki

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.

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