# Volume generated by 2sin(x/2) & sin(x) about y=2 with shells

1. Mar 5, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Find the volume generated by
x=0
x=π
y=2sin(x/2)
y=sin(x)
using the shell method.
2. Relevant equations
x=2arcsin(y/2)
x=arcsin(y)

3. The attempt at a solution

So to setup my integral I got the following values,
I'm pretty sure outer/inner radius is for the shell method so I was thinking a way to get around this would be by setting up two different integrals to get the volumes and then subtracting the smaller one from the larger one.
I'm stuck at finding out the length because for length I have
L=π-x
I'm confused as to which function I should to use here or is this similar to the outer/inner radius where I use them separately for their respective integrals?

2. Mar 5, 2015

### SammyS

Staff Emeritus
(You should give the problem statement directly in your post without regard to what you put in the thread title.)

Your length is incorrect. To help get it correct, sketch the region that is to be rotated about y = 2.

What is the integration variable? If it's y, then x should be in terms of y .

3. Mar 5, 2015

### Potatochip911

Okay after looking at my sketch I think I need to separate this up a lot.
for 0->1 and x=0 to x=pi/2: It appears as though the length is 2arcsin(y/2)-arcsin(y) and then from x=pi/2 to pi length=pi-arcsin(y)
Is this correct?

4. Mar 5, 2015

### Potatochip911

Okay and from y=1 to 2 the length=pi-2arcsin(y/2)
So to get the volume of the solid now I would integrate these lengths with the corresponding radius? What I mean is:
The lengths L=2arcsin(y/2)-arcsin(y) & L=pi-arcsin(y) with Outer Radius=(2-arcsin(y)) and L=pi-2arcsin(y/2) with Inner Radius=(2-2arcsin(y/2))

5. Mar 5, 2015

### SammyS

Staff Emeritus
Almost correct.
arcsin(y) will not give a result greater than π/2, so, to get the x from the y value, for the for the y=sin(x) curve beyond x = π/2, you need x = π - arcsin(y).

Don't forget y from 1 to 2 .

6. Mar 5, 2015

### SammyS

Staff Emeritus
I deleted that post, but not in time.

7. Mar 5, 2015

### Potatochip911

Okay I'm slightly confused, don't I already have x=π-arcsin(y) for the y=sin(x) curve?

8. Mar 5, 2015

### SammyS

Staff Emeritus
What portion of the y = sin(x) curve, are you using the expression,x=π-arcsin(y) for the length?

9. Mar 5, 2015

### Potatochip911

from x=pi/2 to x=pi and y=0 to 1

10. Mar 5, 2015

### SammyS

Staff Emeritus
Right, but arcsin(y) only gives values from pi/2 to 0 for y from 1 to 0. So to get these x values (not the lengths), you need x = π - arcsin(y). Then to get the associated length, take π minus this x value. → length = π - ( π - arcsin(y) ) . The x values here will be between π/2 an π, The lengths will be between π/2 an 0 .

11. Mar 5, 2015

### Potatochip911

Okay thanks I understand what you mean now although I should probably review inverse trigonometric functions. I have been looking at my graph and I think I mixed up the length L=2arcsin(y/2)-arcsin(y) from x=0 to pi/2, should this be L=arcsin(y)-2arcsin(y/2) since x=arcsin(y) is the curve furthest to the right?

12. Mar 5, 2015

### SammyS

Staff Emeritus
Yes.

To check that, try y = .5, for instance. The answer should be positive. (and ... It is.)

13. Mar 5, 2015

### Potatochip911

Okay so now to calculate the volume I am thinking this is a method of doing it:

Upper & Lower limits for the integrals in order: 0-1; 0-1; 1-2
V=2π*∫(arcsin(y)-2arcsin(y/2))*(2-arcsin(y))*dy+2π*∫(π-(π-arcsin(y))*(2-arcsin(y))*dy+2π*∫(2π-2arcsin(y/2)*(2-2arcsin(y/2))*dy

Sorry if this is really hard to read but I can't seem to get latex to work.

Last edited: Mar 5, 2015
14. Mar 6, 2015

### SammyS

Staff Emeritus
I see now that the expressions for radius were wrong. Each should be simply $\ 2 - y\$ .

Also: Notice that π-(π-arcsin(y)) = arcsin(y) .

15. Mar 6, 2015

### BvU

Re TeX:

$\#\#$ V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy $\#\#$

gives

$V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy$

whereas

$$V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy$$

gives displaystyle (bigger int sign, bigger y/2):

$$V = 2\pi \int \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \left (2-\arcsin y\right ) dy$$

Last edited: Mar 6, 2015
16. Mar 6, 2015

### Potatochip911

Whoops, is the volume integral from my previous post correct though if the radiuses are (2-y) or am I setting it up improperly?

17. Mar 6, 2015

### SammyS

Staff Emeritus
Right, (2-y) is what you should use for the radii.

For example, the red in the following should be changed $\displaystyle\ 2\pi \int_0^1 \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \color{red}{\left (2-\arcsin y\right )} dy\$
to $\displaystyle\ 2\pi \int_0^1 \left (\arcsin y - 2\arcsin{y\over 2}\right )\; \color{red}{\left (2- y \right )} dy\$

Likewise in the other integrals .

18. Mar 7, 2015

### Potatochip911

Okay thanks for the help this question was a real hassle considering how much easier it would have been using disks.