Integrals in Polar Coordinates

Date: 04/15/2003 at 17:20:24
From: David
Subject: Integrals in polar coordinates

How can I evaluate an expression like (sin(theta)*(constant)) for 
theta = 0...2*Pi? 

The result is 0, but obviously the volume under the curve can't be 0.

Date: 04/15/2003 at 18:24:14
From: Doctor Douglas
Subject: Re: Integrals in polar coordinates

Hi, David,

Thanks for writing to the Math Forum.

I'm not sure I understand you here.  The area under the curve

  sin(theta*K)

from theta=0 to theta=2*pi is zero only if K is an integer.

Does that help clarify things?

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/15/2003 at 20:07:24
From: David
Subject: Integrals in polar coordinates

Here are more details about the problem:

Triple integration of (x^3 + x*y^2 + x*z^2) dxdydz

Area: 0 <= z <=2, x^2+y^2+z^2 <= 4;

So, in spherical coordinates:

Area: theta(t): 0 <= t <= 2*Pi
      Phi(p): 0 <= p <= Pi/2
      Rho(r): 0 <= r <= 2

And the function to integrate:

(x^3 + x*y^2 + x*z^2) = x*(x^2 + y^2 + z^2)

so, in spherical coordinates

Triple integration = (r*(sin p)*(cos t))*(r^2)*(r^2 * sin p) drdpdt
                   = (r^5)*(sin p)^2 * (cos t) drdpdt 

After I integrate r and p, I get

int (32/3)*(Pi/4)*(cos t) dt

Evaluating this from t = 0 to t = Pi after integration, it gives 0. 
That doesn't seem to make sense... where is my error? Or is 0 the 
real answer?

Date: 04/15/2003 at 22:41:34
From: Doctor Douglas
Subject: Re: Integrals in polar coordinates

Hi again, David.

The region of integration is symmetric in x. And the integrand is odd 
in x (it contains only odd powers of x, such as x and x^3).

Hence the integral is zero by inspection.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/15/2003 at 23:37:02
From: David
Subject: Thank you (Integrals in polar coordinates)

Thanks a lot, Dr. Douglas! I am happy to see that I was not so wrong 
after all. And thanks for answering so fast.

David