Finding Average Value of function

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

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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