Finding Average Value of function

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SUMMARY

The discussion focuses on finding the average value of the function g(x,y) = x^2 + y^2 over the region defined by the equation x^2 + 2xy + 2y^2 - 4y = 8. The user successfully completed the square to transform the region into (x+y)^2 + (y-2)^2 = 12, identifying it as a circle with radius 2√3. However, a critical error was pointed out regarding the nature of the region, which is actually a hyperbola, not a circle. This distinction is crucial for correctly setting up the integration process.

PREREQUISITES
  • Understanding of polar coordinates in integration
  • Knowledge of completing the square in algebra
  • Familiarity with the concept of average value of a function over a region
  • Ability to identify conic sections, specifically hyperbolas and circles
NEXT STEPS
  • Study the properties and equations of hyperbolas
  • Learn how to set up integrals for functions over non-circular regions
  • Explore the method of finding average values of functions in multivariable calculus
  • Review polar coordinate transformations in double integrals
USEFUL FOR

Students studying multivariable calculus, particularly those focusing on integration techniques and the properties of conic sections. This discussion is also beneficial for anyone tackling similar homework problems involving average values and geometric interpretations of functions.

Jimmy21
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Homework Statement



Find the average value of the function g(x,y) = x^2 + y^2 on the region x^2 + 2xy + 2y^2 -4y =8

Homework Equations





The Attempt at a Solution



so far, I complete the square of the region that we want to find average value, x^2 + 2xy + 2y^2 -4y =8. And after completed the square I got, (x+y)^2 + (y-2)^2 = 12. Then I let u = x+y, v = y-2, therefore, i got u^2 + v^2 = 12, which is just a circle with radius 2sqrt3. Then I solve for x and y to plug it into the original function, x = u-v-2, y = v + 2.
After that, I plug it into g(x,y), which I then have, [int][int] (u-v-2)^2 + (v+2)^2, integrate from theta = 0 to 2pi, and r = 0 to 2sqrt3, using polar coordinate. Is the way i did on this problem right so far? I'm not exactly sure of myself.
 
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The region of integration is one-dimensional. Justthe circle, not the interior, so there's no need for a double integral.
 
Good point about the one-dimensionality, but that's not a circle, it's a hyperbola. A circle would never have an "xy" term.
 

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