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### Course: Calculus 2>Unit 4

Lesson 11: Volume: washer method (revolving around x- and y-axes)

# Solid of revolution between two functions (leading up to the washer method)

Finding the volume of a solid of revolution that is defined between two functions. Created by Sal Khan.

## Want to join the conversation?

• Can you rotate a function around a curve?
And how would you do that?
• In principle, you can combine the two original functions and then calculate the volume
circumscribed by that new function.
• Is there a specific reason why they call the method the "washer" method?
• why can't I find the same answer if I use ((x)^1/2-x^2)^2 pie? essentially, Is not it the same areas that sal found in the video? If not, why is that?
• I did a little extra research to understand why this method doesn't work beyond the "you can't do that" answer. It is true that the area in the expression int((x)^1/2-x^2)dx is the same as int(x^1/2)dx - int(x^2)dx as we are familiar with finding area under curves. However, this does not apply to the disc method, as the final volume does not vary proportionally with the distance from the x-axis. It is hard to see from the graph, but becomes intuitive if you imagine the cross sections of nested circles. Here's a helpful diagram: http://i.imgur.com/6VGfXXG.png
• How would it work out if these two equations were revolved around a line not perpendicular to the x or y axis? Such as the shape being revolved around f(x)= 3x +1 or even more complex (if it is even possible) revolved around a parabola such as .5x^2 how would those be calculated?
• In case of y=3x+1: First you would have to use trigonometry to find the Area. The two radi can be found out using Pythagoras because the radius is perpendicular to the rotational axis. From there you would use the disk method.
For rotation around a curve: You would have to find an approximation for the radius using some limit function. Instead of integrating a function to find out the area because the height changes you would integrate the area to find the volume because the radius changes. Keep in mind though, that some values could overlap.
• ....so is this the disk method or the washer method....
• The disk method implies there's no "hole" in the middle: it's a simplified version of the washer method, which implies there IS a "hole".
• Does anyone know if it is possible to do the rotational volume of a solid formula backwards? I have the total cubic area, but need to build a logarithmic function that rotates around the x axis, and is kind of shaped like a carrot.
• When you have a question and they give you formulas in y=f(x), what detail should tip me off that I need to make my y=f(x) functions into x=f(y) functions?
• That depends on how you need to express the radius. For example, f(x) = x^2:
Rotation around the x-axis will give us a radius equal to the fuction value,
Rotation around the y-axis will give us a radius equal to the x-value, so we need an expression for the x-value. Thats why we do the inverse of the function.

So the detail that tips you off are rotation around the y-axis.
• For the volume of the cone inside the "truffle," can we just use the V=1/3*sh (calculating volume for cones)? It's easier than taking the integration of disks.
• Of course you could use the formula for the volume of a right circular cone to do that. In fact, do you see how integral practically gives you (derives) the formula for the volume of a right circular cone?
As for being easier, well, if you don't happen to have the volume formula memorized, then integration is probably faster, and, I suspect, that by the end of this course, you will look at ∫πx²dx and see the answer immediately without doing any calculation as it is a trivially simple integral.