phinah 04 Jun 2020, 17:35
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?
Shell Method | Brilliant Math & Science Wiki
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 & 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 & 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.
Murray 04 Jun 2020, 23:09
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.
You need to be logged in to reply.