Volumes of Revolution Cross Sections

ChemE14
Sun, 02 May 2010 15:22:50 +0000

Part B
(The very first thing you should do is graph the function on your calculator)

1) Find left and right bounds.
This should have already been done from Part A, but we’re starting with Part B. We know the right bound is x=10 (it’s given), and the left bound will be where f(x) and the x-axis intersect. Set f(x)=0, and you get x=1.

2) Determine which method to use (disk or washer).
This is where it helps to have graphed the function. The axis-of-revolution, y=3, is horizontal and it intersects f(x) conveniently at x=10, our right bound of the region R. If it helps, sketch out the graph and shade in the region R. Now, can you see that if we rotated this around y=3, we would have a gap in the middle of the solid? (Also, if the problem had this region rotated around y=0, can you see there would be no gap in the middle of the solid?) This means we must use the washer method.

3) Identify R(x), r(x).
R(x) is the biggest distance from region R (don’t get these two R’s mixed up, just going with what the problem uses, and they call it region R) to the axis of revolution, y=3. Just as region R is bounded on the left and right by x=1 and x=10, region R is bounded on the top and bottom by f(x) and y=0 (the x-axis) respectively, correct? So of these to functions, f(x) and y=0, which is furthest from y=3? Hopefully you can see that for any x-value on the interval [1,10], y=0 is always farther away from y=3. So our big R(x) has a very simple equation

R(x) = [distance from y=3 to y=0] = 3 - 0 = 3

Now onto r(x), the smallest distance from region R to y=3. We can see that the upper bound of region are is closer to y=3 for all x in the interval [1, 10]. We also know that this upper bound is f(x), so now we can write the equation for little r(x):

r(x) = [distance from y=3 to f(x)] = 3 - f(x)

4) Plug into the volume equation.
We know the equation for volume, we have our R(x) and r(x), and we have our bounds for integration, so let’s plug all this in:

V = pi * (int. from a to b) ( [R(x)]^2 - [r(x)]^2 ) dx
V = pi * (int. from 1 to 10) ( [3]^2 - [3 - f(x)]^2 ) dx

5) Solve with a calculator.
Plug this into the calculator using MATH-9 (for TI-83/84), and we get:

V = 212.058

So hopefully that gives you the basic idea / thought process for these problems. Now try part c, but the first thing you have to do now is since it is revolved around a vertical axis, convert EVERYTHING (bounds of integration, f(x), etc.) to be in terms of Y instead of X.