Multivariable Cal, (Polar Coordinates)

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SUMMARY

The discussion focuses on evaluating a multivariable integral by converting Cartesian coordinates to polar coordinates, specifically for the function defined by the equation y = (2x - x^2)^(1/2). The user successfully identifies the region of integration as a quarter circle centered at (1,0) with a radius of 1, but struggles to determine the appropriate bounds for θ and r. The suggested approach includes shifting the center of the circle to the origin using the substitution x' = x - 1, which simplifies the conversion to polar coordinates.

PREREQUISITES
  • Understanding of multivariable calculus concepts, specifically triple integrals.
  • Familiarity with polar coordinates and their application in integration.
  • Knowledge of graphing functions in the xy-plane.
  • Ability to perform substitutions in integrals.
NEXT STEPS
  • Study the process of converting Cartesian coordinates to polar coordinates in multivariable integrals.
  • Learn about setting bounds for integrals in polar coordinates, particularly for circular regions.
  • Explore the method of substitution in integrals to simplify complex regions of integration.
  • Review examples of triple integrals in polar coordinates to reinforce understanding.
USEFUL FOR

Students and educators in multivariable calculus, particularly those seeking to master the conversion between coordinate systems for integration and the evaluation of complex integrals.

wildleaf
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Homework Statement



Evaluate by changining to polar coordinates :
#19 in this picture http://i52.tinypic.com/2ngbt5z.jpg

Homework Equations


x = cos θ
y = sin θ
z = z
r^2 = x^2 + y^2 + z^2

∫∫∫w f(x,y) dxdy
= ∫from θ1 to θ2 ∫from r1 to r2 f(cosθ, sinθ, z) (r* dr dθ)

The Attempt at a Solution


The first thing I did is sketch y = (2x-x^2)^(1/2) in xy plane.

y = (2x-x^2)^(1/2)
y^2 = 2x-x^2
0 = y^2 - 2x + x^2
0 = y^2 + (x^2-2x+1) -1
1 = y^2 + (x-1)^2 ====> circle with radius one, centered around (1,0).

I also graphed y=0, x=1, and x=2. We know that we want the region between x=1 and x=2, and it is in positive y-axis due to y = 0. so we have a quarter of a circle that we need integral. I am stuck here, I do not what the bounds for θ and r are.

I think the bounds for θ is 0 ≤ θ ≤ pi/4 but not too sure.
I have no clue how to find the r for the problem... I know it cannot be 1 ≤ r ≤ 2.

I know how to change the given function into polar but I need help finding the bounds, if someone can help me out, please, thanks in adavance.
 
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You might try first using the substitution x' = x-1 to shift the center of the circle to the origin. Then it's pretty straightforward to change to polar coordinates.
 
Hmm. This seems to be a repeat thread. Nvrm.
 
Last edited:

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