Volumes of Solids of Revolution

Shapes are all around us. Whether big or small, many are symmetrical and have a volume. With math, and some given functions, we can now find this volume through the techniques of visualization and integration. Finding this volume is the second application of integration as the first was finding area between two curves.

When rotating a line around a specified axis, we will always end up with a shape of which an infinite amount of disks can be sliced. If you are not able to find a direction to slice in order to find these cylinders, you have done something wrong.

If you are incapable of visualizing a certain shape, use this link if you like cheating:
bit.ly/makeasolid 


Slice this way:

Not this way:

As for setting up the graph, always be aware of the axis of rotation. If rotating over the x axis, you will have a dx and integrate in terms of x. Likewise with rotating over the y axis, you will have a dy and variables in terms of y. Being aware of this relationship will make your life much easier.

In cases where you only have one function and will be finding the total volume of the solid, the formula is as simple as:

In this formula, the integrand is the radius squared. This radius is the original function given that was then rotated. There is no second function. The dx means you will integrate with respect to x, slicing down the x axis. The bounds a to b represent either where the function intersects the x axis, given numbers, or where the function may intersect vertical lines given in the problem. The depends entirely on the problem and usually is simple to find. 

Once you have set this formula up, simply integrate normally. I will not go into simple integration in this blog. If needed, check out some of the past few blogs. 

The difficulty to these problems increase slightly when a second function is given. This will usually cut out a chunk of of the solid that would have been created by the outermost function. A second function is also useful to finding the bounds. Sometime these bounds are where the two graphs intersect. Other times vertical or horizontal lines are given which makes finding these bounds easier. Remember that with respect to x, the bounds go left to right on the x axis. When integrating with respect to y, the bounds will go bottom to top on the y axis. 

When given two functions, it is best to make a sketch. Once you have done this, you can clearly see which function is the outer function once rotated. These types of problems will create rings, donuts, washers, or however you want to think of it. This is essential for setting up your integral. 

The formula for finding this volume with two functions is as follows:

In this example, f(x) is the outer function and g(x) is the inner. Essentially this finds the volume of one minus the volume of another, giving you the amount remaining.

After this is set up, the integration is no different. It is often useful to multiply out all the terms before integrating for easier math, but you already knew that.

Remember that when the axis of rotation is not the x or y axis, you will need to account for this. You do so by subtracting the axis from each individual function before you do any squaring or integrating. If the outer function is y - 5 and the axis is x = 3, the function you will use is y - 8. This is logical because you are essentially shifting the function as if x = 3 were the y axis. Don't overcomplicate it.

Here is an example of a two function problem in which there is an outer shell and an inner one. We want the volume between the two.

Find the volume of the solid generated by revolving the region bounded by y = x ² + 2 and y = x + 4 about the x‐axis.

Sketch.

First find where they intersect. These will be the bounds.

Rotate.

                  

Set up and solve.

Integration Applications

This week we began Integral Applications, our fifth and final unit of AP Calculus. Integral Application is used to find the area bound by multiple functions, with respect to either X or Y.

When finding the area bound by two curves with respect to X, use this formula:

Example:

And when finding the area bound by two curves with respect to Y, use this one:

Example:

Tips:

• Always start by graphing the given functions if a graph isn't given.
• When the bounds are not explicitly stated, you can set the functions equal to each other and solve   for the values which will be your bounds.
• When there is no "bottom" function, you must integrate with respect to Y
• When the given functions switch positions, the integrals should be broken up so that the function that is on top acts as f(x).

The Fundamental Theorems of Calculus

The Fundamental Theorem is what we've been working up to all year. It uses derivatives, anti derivatives, and integrals to find the area beneath a graph. It states:
If f is a continuous function on the closed interval [a,b] and F is the anti derivative of f, then:

This means, in cases where we cannot find the integral of the function easily, we can use the anti derivative between two points and end up with the same answer.
The Fundamental theorem leads into the Mean Value theorem for integrals. This is another method which can be used to find the integral of a function. On any graph with the domain [a,b] there are points on the y axis, [f(a),f(b)] There must be a point between them on the graph, called f(c). Using f(c) and the difference between b and a to make a rectangle, we are able to find the integral of the function once again. There is an image of my personal notes, the important part is the graph that shows where f(c) would go.  and  are the two important functions we use from this theorem. f(c) is found by averaging f(a) and f(b).
We also covered the Second Fundamental Theorem of Calculus on Friday. The second theorem involves taking the derivative of a function defined by an integral. The format of the functions change in this theorem:
 the integral is now an equation, and we have two variables. This theorem focuses on the x in the equation. We can use this function to find the derivative of f(x).
The notes below are from class:

They explain how the second theorem is used using a sine equation for an example. We will generally not be concerned about the second variable, as it only changes the line on the graph and not what we are looking to find. The equation wants to find f(2π) so: 

This equation now looks like something we are more used to. The equation is solved like a normal integral, and the answer is 0.
In the second theorem, the first number in the integral, the "a" value, must always be a number. The second one must always be a single variable.
In class, we also found out what to do when the upper bound is not a single x. You need to find the derivative of the function, and use the first theorem to finish solving.

The last thing we looked at was what to do if the bounds of the function were two variables, rather than a number and a variable.

We can use a constant, called a in the notes, to divide the two sides of the bounds. Then, since a constant is needed in the bottom, one of the functions can be flipped, so they both run from a to x. The flipped function only needs to be multiplied by -1 and the two are added back together to find an answer for the specific problem.