Nagisa Maths Thread

[Nagisa]

the curves and generate an area that is rotated about the line

find the volume of the generated shape

edit: the area between the curves is the area generated

[BubbleWrapMan]

Translate it 1 unit to the left, then rotate it about the y-axis

[Nagisa]

will that make it any easier? as far as algebra

[BubbleWrapMan]

Easier than what? I can't think of another way to do the question.

[Nagisa]

doin volumes with integrals

[BubbleWrapMan]

You still integrate, but the formula only works for rotation about either axis, so you need to move it from x = 1 to x = 0 so you can use the integral formula.

[Planck's constant]

ClimbTooHigh, stellar maths study scores you got there

[FlorianK]

Ok so you could take the integrals with respect to y, but lets find the inverse function and take the integrals with respect to x. But thats just cause I like solids of revolution over the x-axis more than y-axis, you can do it either way.

So we have:
x=3
and
y=2+sqrt(4-x) and y=2-sqrt(4-x)
So now we translate it all 1 unit down

We get:

x=3
and
y=1+sqrt(4-x) and 1-sqrt(4-x)

The are enclosed by the 3 graphs is still the same as in your question(see properties of inverse functions)

Snow we split the Area into 2 areas
The one under the 'upper' square-function and the one under the 'lower' square-function

If we rotate the Area of the 'upper'square-function we get exactly the Volume we want +  the Volume generated by rotating the 'lower'-square-function

The to find the Volume generated by the square-functions we add up all the infinitly many circels generated. The function for the area of a circle of a certain x value is
C(x)=pi*[f(x)]² since f(x) is the radius and A=pi*r²

So now we just take the integral of the circle-function from 3 to 4.

So for the 'upper'-square-function it is



For the 'lower'-square-function it is



So the volume is 8pi/3

Hope I didn't do a mistake.

ClimbTooHigh can you check please?

[BubbleWrapMan]

Looks right.

[TrueTears]

This volume can be found very easily by the following computation:



Why?

[Phy124]

Didn't do spesh so not sure if this is on the course, but how I would have done it using the shell method.

That is

Where in this case is the upper function (translated 1 unit left for the reasons stated in previous posts) minus the lower function over the domain ,, which they have an enclosed area i.e.



edit:

This volume can be found very easily by the following computation:



Why?

Oh yes, you can just multiply the function by (x-1) rather than translating the function left 1, of course

[TrueTears]

Didn't do spesh so not sure if this is on the course, but how I would have done it using the shell method.

That is

Where in this case is the upper function (translated 1 unit left for the reasons stated in previous posts) minus the lower function over the domain ,, which they have an enclosed area i.e.

Correct, using cylindrical shells is what I was referring to, however is there really a need to translate?

Makes the algebra harder, in fact whenever you spin around a different vertical or horizontal line, there is never a need to translate, all you have to do subtract or add the radius/height on.

EDIT: Just saw your edit, yup that's correct Clifford

[Nagisa]

This volume can be found very easily by the following computation:



Why?

this was my method also but couldnt quite get it to work and i didnt know why lol. I had 4 as the int terminal and radius as (1 - x).

[TrueTears]

lol hsc maths is pretty cool

[FlorianK]

TT, why do you destroy my working out with a line ?