Question: What is the volume of the revolution bounded by the curves of y=4-x^2 , y=x, and x=0 and is revolved about the vertical axis.

First, I had found the points of intersection to get the limits and I got -2.5616 and 1.5616. And then I plug it in the shell method formula and I got the lower limit of -2.5616 to upper limit of 1.5616 of the integral of x((4-x^2)-x)dx times all of that by 2pi. I was wondering if I did this all wrong ?

I guess the decimal values are ok, but I tend to prefer the exact values of (-1+√17)/2 and (-1-√17)/2. They are a bit cumbersome to work with by hand, but with the wealth of online web sites to do the calculations, that should be no obstacle.

Your limits of integration are wrong, since x=0 is one boundary. Including area left of the y-axis would involve overlap during the revolution.

So, using shells of thickness dx, the volume

v = ∫[0,(-1+√17)/2] 2πrh dx
where r=x and h=(4-x^2)-x
v = ∫[0,(-1+√17)/2] 2πx(4-x^2-x) dx

Your integrand looks good.

Thanks a bunch Steve!

Based on your approach, it seems like you are using the method of cylindrical shells to calculate the volume of revolution. However, there are a few issues with your setup.

First, to find the correct limits of integration, you need to find the points of intersection between the curves. In this case, you correctly found the points to be -2.5616 and 1.5616. However, it seems like you have only considered one of the points as the lower limit and the other as the upper limit.

To use the cylindrical shells method, you need to split the area enclosed by the curves into two parts, each having a different radius of revolution. One part will be below the intersection point (-2.5616) and the other part will be above the intersection point (1.5616).

Additionally, when using the cylindrical shells method, the integral should be set up with respect to the variable that is the axis of revolution, which is usually either x or y. In this case, since the axis of revolution is the vertical axis, you'll want to integrate with respect to x.

Here's how you can correctly set up the integral using the cylindrical shells method:

1. Find the points of intersection between the curves:
Set y = 4 - x^2 equal to y = x to find the intersection point:
4 - x^2 = x
Rearrange and solve for x:
x^2 + x - 4 = 0
Using the quadratic formula, we get:
x ≈ -2.5616 and x ≈ 1.5616

2. Split the area enclosed by the curves into two parts:
The part below the intersection point (-2.5616) has a radius of revolution given by y = 4 - x^2.
The part above the intersection point (1.5616) has a radius of revolution given by y = x.

3. Set up the integral:
The integral will be with respect to x, as the axis of revolution is vertical.
The limits of integration will be from -2.5616 to 1.5616.

The integrand will be 2πrh, where r is the radius of revolution and h is the height of each cylindrical shell.

For the part below the intersection point (radius: 4 - x^2):
The height of each cylindrical shell is dx.
The radius of revolution is given by r = 4 - x^2.
The integral for this part is ∫[from -2.5616 to 1.5616] of 2π(4 - x^2)dx.

For the part above the intersection point (radius: x):
The height of each cylindrical shell is dx.
The radius of revolution is given by r = x.
The integral for this part is ∫[from -2.5616 to 1.5616] of 2πxdx.

4. Add up the two integrals to get the total volume of revolution:
Volume = ∫[from -2.5616 to 1.5616] of 2π(4 - x^2)dx + ∫[from -2.5616 to 1.5616] of 2πxdx.

By correctly setting up these integrals, you can calculate the volume of revolution using the cylindrical shells method.