Integrals in Polar CoordinatesDate: 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 |
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