Originally Posted by

**eddie2042** Ok here's when to use the methods.

1) Use disks when the function is defined in terms of a variable who's line of rotation is PARALLEL to the axis of that variable.

So for example; say you have a function that is defined in terms of x (i.e. y = 1 + sin (x)) and you're revolving it around a line (i.e. y = -1). notice, that the line y = -1 is parrelel to the x-axis. In this case, you use the method of disks which is defined as:

$\displaystyle V = \pi\int_a^b(f(x))^2dx$

where $\displaystyle f(x)$ represents the RADIUS of the cross-sectional circle of the rotation. Sometimes, it may not be just $\displaystyle f(x)$. In this case, we're revolving it around y = -1, so the radius isn't going to be $\displaystyle f(x)$, it's going to be $\displaystyle f(x) +1$ (think about it).

Another example. Say you have a function defined in terms of y (i.e. x = y^2 - 3y) and you want to rotate it around x = 4. This line is parallel to the y-axis (which is the variable the function is in terms of).

2) The method of washers is simply an extension of the method of disk. It's when you have two functions and you're trying to rotate the volume enclosed/underneath them.:

$\displaystyle V = \pi\int_a^bf(x)^2 - g(x)^2dx$

over the interval in which the volume produced by $\displaystyle f(x)$ is greater than the volume produced by $\displaystyle g(x)$.

DO NOT make the mistake of writing the integrand as

$\displaystyle (f(x) - g(x))^2$ << this is simply wrong (try figuring out why).

3) The method of cylindrical shells is used when the line you're rotating about is PERPENDICULAR to the axis of the variable the function is defined as. So, if you got a function, y = x^2, and you want to rotate it about y = 0 (the y axis), you can use the method of cylindrical shells:

$\displaystyle V = 2\pi\int_a^bx(f(x))dx$

Where the $\displaystyle x$ represents the radius of the cylinder. Once again, it may not always be $\displaystyle x$ on its own. It is if you're rotating it about the central axis, but if you're rotating, let's say, y = sqrt(x), from [4,9] about the line y = -1, then the radius will be (x+1) and that's what you put in place of the original x. The $\displaystyle f(x)$ represents the height of the cylinder. You may have to change it if you're revolving the area enclosed by two graphs, so you'll have to put i.e.$\displaystyle f(x) - g(x)$ in place of the original $\displaystyle f(x)$

OK SO... to answer your original question.. what do we for y = 1 + sin (x)? Once again, use the method of disks.

$\displaystyle V = \pi\int_0^\pi(1 + sin(x) (+1))^2dx$ << the +1 is there because the radius isn't just f(x) (remember?)

$\displaystyle V = \pi\int_0^\pi(2 + sin(x))^2dx$

$\displaystyle V = \pi\int_0^\pi sin^2(x) + 4sin(x) + 4dx$

here's where your hint comes in handy, though i will write it slightly differently..

$\displaystyle V = \pi\left[-\frac{sin(2x) - 2x}{4} - 4cos(x) + 4x\right]_0^\pi$

$\displaystyle V = \pi\left[(\frac{\pi}{2} + 4 + 4\pi)-(-4)\right]$

$\displaystyle V = \frac{9\pi^2 + 16\pi}{2}$

I could be wrong...