<p>Can someone explain a general method on how to do these? My teacher never formally taught them so I’m a little lost.</p>
<p>This is a very quick explanation (though it’s been discussed in more detail in other AB/BC threads on here).</p>
<p>Rotational Volume (I’ve talked more about this on at least one other BC thread, so you can check my recent posts to find those links):
Disk method
V = pi * integral from a to b [R(x)]^2 dx, where R is the distance from the function to the axis of revolution</p>
<p>Washer method
V = pi * integral from a to b [ [R(x)]^2 - [r(x)]^2 ] dx, where R is the bigger distance from the axis of revolution and r is the smaller distance from the axis of revolution</p>
<p>Volume by Cross-Sections:
V = [integral from a to b] A(x)dx, where A is the cross-sectional area. It’s pretty easy once you get the hang of it. The problem will tell you the shape of the cross-sections (semicircle, square, rectangle with a certain height, triangle, equilateral triangle, etc.) and, if applicable, the height of those cross-sections (the height can be constant or in terms of x). The base of the cross-sections is the distance from the upper-y-bound (this is usually the function) to the lower-y-bound (this is usually a horizontal line).</p>
<p>Example: f(x) = x^2, where region R is bounded by x=3, x=1, and y=0. Find volume with semicircle cross-sections coming out perpendicular to the x-axis.</p>
<p>1) Identify your equation for cross-sectional area
What is the area of a semicircle? Half the area of circle, so
A = pi * r^2 / 2
So now we must identify ‘r’, or the radius of the semicircle. You see that the base of each semicircle is the distance from f(x) to y=0, right? So the base = f(x) - 0 = f(x). However, this base represents the diameter of the semicircle - for the equation, we need the radius, so</p>
<p>r = d/2 = f(x)/2</p>
<p>Now plug this back into the equation to get</p>
<p>A(x) = pi * [f(x)/2]^2 / 2 = (pi/2) * [f(x)]^2 / 4 = (pi/8) * [f(x)]^2</p>
<p>All you have to do now is integrate this from x=1 to x=3, and you have your answer.</p>
<p>Your book may have a list of specific equations for each cross-sectional shape, and while you can memorize them, I personally think it’s just easier to derive the cross-sectional volume equations yourself.</p>
<p>Umm, the key to doing these, in my opinion, is to make a sketch graph of the cross section. Identify the bounded region, lightly shade it, and draw the axis of revolution. Draw in a rectangle which will help you determine dx or dy, disk/washer or shell, and the limits of integration. </p>
<p>With disk/washer, you need to identify big R and little r. With shell, you need p and h. A rectangle that is vertical requires integration with respect to x (dx) and a horizontal rectangle necessitates dy. You determine your limits based on how far along the axis the rectangles move. If it’s a dx problem, it’s the left and right x-values. If it’s a dy problem, it’s the y-value at the bottom and top.</p>
<p>I’m sure someone will come in and explain this better; I’m not great at teaching volume of solids of revolution.</p>
<p>These questions sometimes appear as one of the parts on a free response for area/volume problems. The pattern seems to be that you find the area of the region in the first part of the free response and then one of the other parts of the problem is to find the volume with a known cross section. You have to use the area that you find in the first part in the integral you use in the cross section part of the question. In order to find the volume with the known cross section, it depends on what type of cross section the problem is asking for. If the problem states that the known cross section is: </p>
<p>A square - you take the integral of the following: ( base^2 ) </p>
<p>An equilateral triangle - you take the integral of the following: ( base^2 * (√3)/4 ) </p>
<p>A semicircle - you take the integral of the following: ( radius^2 * pi/8 )</p>
<p>Make sure when you take the integral that the bounds are the same as the bounds for the area you found in the first part of the problem. I hope this helps.</p>
<p>^^^
Made a typo for square and equilateral triangle. “Area” should be replaced with “base.”</p>
<p>And semicircle is the integral of (radius ^ 2 * pi/8), as I showed above.</p>
<p>Do not listen to @byubound. </p>
<p>For a square you do not integrate (area^2). You integrate (area) or (side^2).</p>
<p>Triangles depend on the problem.</p>
<p>Semicircles are (radius ^2 * pi/8).</p>
<p>Thanks for editing your post. ;)</p>
<p>Thanks for catching that 
I fixed it. Sorry for the confusion. I think I confused area with base because I was trying to explain that you take the area you found in the first part to use as the base in the cross section part of the problem. Usually it is stated that the region from the first part is the base.</p>
<p>Beat you to it! (EDIT: This was addressed to NewAccount - I’m too slow)</p>
<p>This is why I really don’t like memorizing the final formulas. If you put pi/2 instead of pi/8, or sqrt(3)/2 instead of sqrt(3)/4 (how did you make the sqrt sign?), you get it all wrong. But if you work it from the beginning, your final formula is usually right. (However, if you have a greater chance of messing up by working the area equation out instead of memorizing it, then by all means memorize. Just do what’s easiest for you)</p>
<p>I just copied the square root off of a math page. I looked those up in PR and they haven’t steered me wrong on any FRQs I’ve done. My teacher said it would be easier to memorize those than to try deriving them; have I been led astray?</p>
<p>No, you haven’t been led astray. It’s just easier for ME to derive them (I don’t like memorization too much - for example, on an equilateral triangle I would rather find the height by splitting the thing in two and doing 30-60-90 rule, instead of memorizing A=sqrt(3)/4 * b^2 - don’t ask why). If it’s easier for you to just memorize the equations, then that’s perfectly fine. Some people in my class relied on the equations so much that they forgot the basic concept of the whole thing, and on one problem we worked, the equation for the cross-sectional area was GIVEN (something like A(x) = 4 + 3x/2), but they were lost because they were wondering what shape it was, they didn’t think they had an equation. So as long as you know the basic idea behind the whole thing, then yes, go ahead and memorize it.</p>
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<p>I’m pretty sure you get the idea, but just wanted to point out to those who don’t that this phrase isn’t completely right. In those equations, “base” represents the length of the line from the upper bound to lower bound, not the area of the whole region. What you’re doing when you’re integrating this area function is adding up an infinite amount of cross-sections. I think of it as adding up a lot of very thin “slices” (the cross-sections) to get the total volume of the solid.</p>
<p>If that confused anyone, sorry, I have trouble explaining it. My main point to get across was that “b” (the base) is not the area of the region but a distance from the top to bottom of that region.</p>
<p>Can anyone explain how to exactly solve solid of revolution problems that rotate over something other than the y and x axises? Something like x=-2, y=3, etc.</p>
<p>If the axis-of-revolution is horizontal (y=), then the function has to be in terms of x and you integrate with respect to x.</p>
<p>If the axis-of-revolution is vertical (x=), then the function has to be in terms of y and you integrate with respect to y.</p>
<p>Then all you have to do is identify big R and little r (for washer method). Big R is the larger distance from the region to the axis-of-revolution, and little r is the smaller distance to the axis-of-revolution.</p>
<p>Yeah, but how exactly do you subtract from the integrands? Like, say, a function is rotated about the line x=5. What exactly should I do to translate the function?</p>
<p>Also, a rather off-topic question: sometimes, you have to find critical points of a very complex function. How exactly would I go about doing that? Even a calculator fails to give me the critical points of functions like x+4e^(sin4x). I just get a blank list. Is there a specific program or trick to do these?</p>
<p>For the critical points question, just take the derivative and graph it in your calculator. Whenever the graph of f’ crosses the x-axis, that’s a critical point (rather, the x-value). It’s a pretty simple derivative. If the derivative never crosses the x-axis, there are no critical points.</p>
<p>I’m not sure what you mean by a “blank list” when you look for the critical points.</p>
<p>Washer method
V = pi * integral from a to b [ [R(x)]^2 - [r(x)]^2 ] dx, where R is the bigger distance from the axis of revolution and r is the smaller distance from the axis of revolution</p>
<p>That’s the equation in terms of x for finding the volume rotated around an axis. So if you’re rotating around a vertical axis (x=5), then change all the x’s above to y’s. If you provide a specific example, I’ll be glad to work it out step by step.</p>
<p>I’ll help out a little bit too. When you do revolution and you don’t know whether to use dy or dx, just look at what axis you’re revolving the function around. If it’s x = something, use dy. If it’s y = something, use dx. Don’t get mixed up on the ‘if it’s revolved around the y axis, use dx.’ Y axis is x = 0, so you have to use dy. Same with the x axis, use dx for that.</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
Number one, parts b and c. </p>
<p>Thanks for the help guys.</p>
<p>Part B
(The very first thing you should do is graph the function on your calculator)</p>
<p>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.</p>
<p>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.</p>
<p>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</p>
<p>R(x) = [distance from y=3 to y=0] = 3 - 0 = 3</p>
<p>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):</p>
<p>r(x) = [distance from y=3 to f(x)] = 3 - f(x)</p>
<p>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:</p>
<p>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</p>
<p>5) Solve with a calculator.
Plug this into the calculator using MATH-9 (for TI-83/84), and we get:</p>
<p>V = 212.058</p>
<p>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.</p>