# Solid of Revolution of an arcsin graph

1. Oct 23, 2011

### e to the i pi

1. Consider the function f(x) = arcsin(2x^2 - 1).
Write down, but do not attempt to solve, a definite integral in terms of y, which when evaluated will give the volume of the solid of revolution formed by rotating the graph about the y-axis.

2. Endpoints are at (-1, pi/2) and (1, pi/2).
x-intercepts are at (-1/sqrt(2), 0) and (1/sqrt(2), 0).
y-intercept is at (0, -pi/2).
The equation looks like a "bowl" shape.

3. I know that there are two main methods: the disc method and the shell method. Unfortunately, I don't know how to apply them very well. I'm used to having functions only in one quadrant, but this function is in all 4 quadrants at once!
My best attempt would be something like this:
Trying to put it in the form: 2pi*r*h
Integral from -1 to 1 of: 2pi * arcsin(2x^2 - 1) * dy

2. Oct 23, 2011

### HallsofIvy

You need of course, to write that integrand in terms of the variable of integration, y.

If y= arcsin(2x^2- 1, then 2x^2- 1= sin(y), 2x^2= 1+ sin(y), x= sqrt{(1+ sin(y)/2). By symmetry, it is sufficent to find the volume generated by rotating the graph in the first quadrant around the y- axis and then multiply by two.

3. Oct 30, 2011

### e to the i pi

So once I have it in the form x = sqrt((sin(y) + 1)/2), do I then have to change the numbers on the top and bottom of the integral? So instead of going from -1 to 1, I am now going from -pi/2 to +pi/2? And by doing that, I am now wrapping it around the x-axis? So I can change it to become:
y = sqrt((sin(x) + 1)/2) from -pi/2 to pi/2 around the x-axis?
If that is correct, what do I do from then on to create the definite integral?

4. Oct 30, 2011

### e to the i pi

You said I should cut it in half and then calculate half of that. Does that mean I am only going from 0 to +pi/2 or from -pi/2 to 0, rather than the full way?
So would the integral be like this:
Integral from 0 to +pi/2 of: (sqrt((sin(x) + 1)/2)) * dx

5. Nov 5, 2011

### e to the i pi

Please someone help me! It's been literally weeks and I still can't solve this problem :(

6. Nov 7, 2011

### e to the i pi

Okay, my new attempt at the solution is to use the formula:
V = Integral from a to b of [f(y)]^2 * dy
By transposition we get:
x = sqrt((sin(y) + 1) / 2)
Since x = f(y), then [f(y)]^2 = (sin(y) + 1) / 2
Now I thought that since an arcsin graph is symmetrical, I can just simply calculate it from -pi/2 to pi/2 since when it rotates it's going to cover the other half anyway.
So the definite integral goes from -pi/2 to pi/2
V = Integral from -pi/2 to pi/2 of (sin(y) + 1) / 2 * dy
Is that correct?

7. Nov 13, 2011

### e to the i pi

Please tell me if I'm doing it right or wrong because I am still confused!

8. Nov 16, 2011

### e to the i pi

9. Nov 16, 2011

### jackmell

I think what happens is people see a list of answers and just assume it's answered. Just use volume by shells along the y-axis where the volume of the i'th shell is just $\pi r^2 dy$ right? but keep in mind, wish to have the radius always positive so when you solve for the radius in terms of x, you get:

$$x^2=1/2 \sin(y)+1$$

but y goes from -pi/2 to pi/2 and the negative values would give a negative value of the sine. So once you get that straight, you can write the volume as:

$$V=\pi \int_{-\pi/2}^0 (1/2\sin(-y)+1)dy+\pi \int_0^{\pi/2}(1/2\sin(y)+1)dy$$

think so anyway. You go through it and make sure everything makes sense to you.