The Integration problem that's still giving me a headache (a week later)

[Sombra]

Last week (or longer) I asked how to integrate the following rotated 360 degrees about the x-axis (and I still can't figure it out):
y= [(1-(x-3)^2)/9)]^(1/2) + 2

When I applied the boundaries of 0 to 6, I got 23.2 cm^3, but I don't think its the boundaries that are my problem, I think it is the integration. Help! I need this for tomorrow. Thanks!

 

[HallsofIvy]

First, it makes no sense to "integrate the following rotated 360 degrees"! You integrate a function but you can't rotate a function- you rotate a geometric object!

My best guess is that you want to find the volume of the figure formed by rotating the graph of y= [1-(x-3)/9]^1/2+ 2 around the x axis. Doing that, each cross section is circle of radius y so the each cross section has area πy² and so the function you want to integrate πy²dx. That should not be hard.


If y= [1- (x-3)²/9]^1/2+ 2 then (y-2)²= 1- (x-3)²/9 so (x-3)²/9+ (y-2)²= 1. That's an ellipse with center at (3,2) and semi-axes of length 1 (in the y direction) and 3 (in the x direction). It doesn't cross the x-axis but since you specify y= [1- (x-3)²/9]^1/2+ 2, I GUESS you mean to drop straight lines from the largest and smallest x values, 3-3= 0 and 3+3= 6, the values you mention so I guess I am interpreting this correctly. If that's the case then your integral is
$$\pi\int_2^4 y^2 dx= \pi\int_2^4(1-\frac{(x-3)^2}{9}+ 4(1-\frac{(x-3)^2}{9})^{1/2}+ 4)dx$$
You should be able to integrate all except that middle term, with the 1/2 power, easily.

To integrate $4(1- \frac{(x-3)^2}{9})^{1/2}$, use a trig substitution.
let [tex]sin(\theta)= \frac{x-3}{3}[/itex]. Then $cos(\theta)d\theta= \frac{1}{3}dx$ so that $3cos(\theta)d\theta= dx$ and $(1-\frac{(x-3)^2}{9})^{1/2}= (1- sin^2(\theta))^{1/2}= cos(\theta)$.[/tex]

 

[thepatientmental]

I can understand your frustration with this integration problem. It can definitely be challenging to integrate a function that has been rotated about an axis, especially when it involves a square root.

First, let's focus on the integration itself. One approach you can take is to use the substitution method. Let u = (1-(x-3)^2)/9. This will simplify the function and make it easier to integrate. Then, after integrating, you can substitute back in for x.

Another approach is to use the disk or washer method. In this case, you would need to split the function into two parts - the upper half and the lower half of the rotated shape. Then, you can use the formula for finding the volume of a solid of revolution to integrate each part separately and then add them together.

As for the boundaries, it's important to make sure they are correct for the problem. Double check to see if the function is being rotated about the x-axis or the y-axis. Also, make sure that the boundaries are in terms of x and not y.

I would recommend seeking help from a tutor or classmate if you are still struggling with this problem. It's always helpful to have someone explain it to you and walk you through the steps. Don't stress too much, with some practice and guidance, you will be able to solve this type of integration problem. Good luck!