I'm trying to take a simple equation: y = 2x and integrate it using the polar co-ordinate system between x = 0 and x = 1.
In the cartesian system I can take the double integral 2xy dx dy with the limits from 0 to 2x and 0 to 1 for dx dy respectively.
Converting to polar coordinates I know I can use x = rcos(theta) and y = rsin(theta) and dx dy = rdrd(theta)
this gives me the double integral of 2r^3cos(theta)sin(theta) dr d(theta) with limits 0 to sec(theta) and 0 and 3pi/8 for dr and d(theta) respectively.
I used sec(theta) because r = a/cos(theta), at x=0 r = 0 and x = 1 it will therefore be sec(theta). 3pi/8 is the angle from x to r.
when i calculate this integral however I get an answer of about 1.46
the integral of y=2x between 0 and 1 is 1. Seeing as using polar coordinates would take an arc I would expect the area to thus be less than 1.
Am I missing something here?
In the cartesian system I can take the double integral 2xy dx dy with the limits from 0 to 2x and 0 to 1 for dx dy respectively.
Converting to polar coordinates I know I can use x = rcos(theta) and y = rsin(theta) and dx dy = rdrd(theta)
this gives me the double integral of 2r^3cos(theta)sin(theta) dr d(theta) with limits 0 to sec(theta) and 0 and 3pi/8 for dr and d(theta) respectively.
I used sec(theta) because r = a/cos(theta), at x=0 r = 0 and x = 1 it will therefore be sec(theta). 3pi/8 is the angle from x to r.
when i calculate this integral however I get an answer of about 1.46
the integral of y=2x between 0 and 1 is 1. Seeing as using polar coordinates would take an arc I would expect the area to thus be less than 1.
Am I missing something here?