Washers and Discs, Graphs and Symmetry
Date: 2/20/96 at 20:2:7 From: Anonymous Subject: Washers and Discs Dear Dr. Math, I am a Calculus student living in Augsburg, Germany. Our class has encountered a problem in solving one of the problems in our book. The problem is as follows: Use the washer method to find the volumes of the solid generated by revolving the regions bounded by the lines and curves about the y-axis. 30. The semicircle X=(25-y^2)^(1/2) and the line X=4 We figured out how to do it by working it and looking in the answer book; however, the answer book has 2pi in front of the integral symbol. We believe it should only be pi. Please Explain? Thanks, Ara Donabedian, Anna Schmidt
Date: 9/25/96 at 15:6:48 From: Doctor Jerry Subject: Re: Washers and Discs Maybe you forgot the bottom half of the area. That would explain the missing factor of 2. The piece of the semicircle to the right of the line x=4 is symmetric about the x-axis. Usually it is good practice to use symmetry in solving area and volume problems. In case it is useful, here is an outline of my solution. The area lies to the right of the line x=4 and to the left of the semicircle x=(25-y^2)^(1/2). For the washer method, the "elements" will be horizontal. I would write dV= pi*x2^2*dy-pi*x1^2*dy where x2=(25-y^2)^(1/2) and x1=4. The expression for dV simplifies to dV=pi*(25-y^2-16)*dy. When this is integrated for y=-3 to y=3 (or from y=0 and y=3 and then multiply the result by two), I obtain V=36*pi. -Doctor Jerry The Math Forum
Dear Ara and Anna: Hello! I'm glad you are double checking the answers the book gave you. Sometimes books can be wrong, and anyway, it is a good idea to have a firm understanding of the reasons for an answer to a problem. I'm not sure how the book set up the problem, but I'm guessing that perhaps the book utilized the fact that the graph you are rotating about the y-axis is symmetric about the x-axis. So, let me say a little about symmetry of graphs and how we can use symmetry to make calculations easier. If you draw the graph described in the problem, you notice that it has the same area above the x-axis as it does below the x-axis. Thus, instead of figuring out what the area of the whole graph is directly, we can calculate the area of the graph above the x-axis. We then multiply this area by 2 to get the total area of the region described, since the area of the graph below the x-axis is the same as the area of the graph above the x-axis. In the preceding paragraph I was talking about the area of the given graph. You can probably see that this concept can be applied to finding the volume of a surface of revolution. If we are rotating about the y-axis a given area that is symmetric about the x-axis, then instead of calculating the volume in the standard way, we can calculate the volume of the surface of revolution we get when we rotate the upper portion of the area about the y-axis. To get the total volume, we need only multiply by two. In his response to you, Dr. Jerry goes through the calculations, so I won't repeat that here. You might wonder why we would use this symmetry when we could have gotten the answer using standard techniques. Well, sometimes the integrals are easier to evaluate when we use symmetry (this is because often one of the limits of the integral is 0 and things are often easier to deal with when they involve a 0!). However, standard techniques of integration would also work on problems such as these. You should note that here symmetry made our problem easier to calculate. Symmetry is often useful in making calculations easier in lots of different fields of math - not just calculus. In addition, mathematicians admire and study symmetry for its aesthetic value. If you would like to know more about how symmetry comes into play in mathematics, please do write back to us! I hope this helps. -Doctor Sydney, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.