Expressing the limits of integration for radius in polar coordinates

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SUMMARY

The discussion focuses on integrating a function bounded by the circle defined by the equation x² + y² = 6y in polar coordinates. The user seeks to determine the appropriate limits of integration for the radius (r) in the double integral ∫∫f(x)r dr dθ. A key insight provided is to express the equation in polar coordinates, resulting in a simpler form that directly yields the bounds for r. The suggestion to shift the origin is deemed unnecessary, as the polar equation can be derived without this adjustment.

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  • Understanding of polar coordinates and their conversion from Cartesian coordinates
  • Familiarity with double integrals and their applications
  • Knowledge of the equation of a circle in Cartesian coordinates
  • Basic calculus concepts, particularly integration techniques
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i'm trying to integrate some some function bounded by the x-y domain of x2+y2=6y

which is a circle on the x-y plane shifted upward where the outer part of the circle is 6.

i'm trying to integrate a double integral.. ∫∫f(x)rdrdθ

i don't know how to express my limits of integration for r.

the only thing i can think of is going from 3 to 6 but then this means that i have a radius three units long going to a radius of 6 units long. what I'm looking for is a radius 0 units long to radius 3 units long, i just need this shifted up 3 units. how do i express my limits of integration for the r component?

thanks
 
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You can choose the origin of he polar system of coordinates at (0,3), and change the variable y to u=y-3 in the integrand.

ehild
 
You don't need to shift the origin. Write the equation ##x^2+y^2=6y## in polar coordinates to get a simple polar equation in the form ##r = f(\theta)##, which will give you the ##r## bounds.
 
Last edited:

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