Why Does dxdy = rdrd(theta)? Why Not Just drd(theta)?
Date: 12/29/2009 at 14:48:08 From: scott Subject: why dxdy = rdrd(theta) from x=rcos(theta) and y=rsin(theta) In transforming from rectangular form to polar form, I know that x = rcos(theta) and y = rsin(theta) But why does dxdy = rdrd(theta)? Why the additional r term? Why isn't dxdy equal to just drd(theta)? I have always memorized this without understanding it. I would like to know how to prove this. Thanks in advance for your help.
Date: 12/29/2009 at 17:50:21 From: Doctor Jerry Subject: Re: why dxdy = rdrd(theta) from x=rcos(theta) and y=rsin (theta) Hello Scott, Thanks for writing to Dr. Math. Look at this figure: It shows the area corresponding to an increase in theta of d(theta) and an increase in r of dr. The small figure with sides of dr and r*d (theta) is very nearly a rectangle, and has area r*dr*d(theta). Please feel free to write back if you have questions. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
Date: 12/30/2009 at 12:00:11 From: scott Subject: Thank you (why dxdy = rdrd(theta) from x=rcos(theta) and y=rsin(theta)) Thank you very much. You guys are awesome. I really appreciate it.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.