Question

Now, substitute x in terms of y into the formula for F: F= 1/Area(D) ∫0^15 ∫0^y^2 +12(x^2 +y^2) dxdy Now, you need to evaluate this double integral to find the average square distance F.

          Now, substitute x in terms of y into the formula for F:
F= 1/Area(D) ∫0^15 ∫0^y^2 +12(x^2 +y^2) dxdy
Now, you need to evaluate this double integral to find the average square distance F.
        

Added by Lauren D.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Now, substitute x in terms of y into the formula for F: F= 1/Area(D) ∫0^15 ∫0^y^2 +12(x^2 +y^2) dxdy Now, you need to evaluate this double integral to find the average square distance F.
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Transcript

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00:01 So in this question, we want to find the average value of the function f of xy equals x squared plus y squared over the square having vertices at 0 0, 02, 22, and 200.
00:17 So how do we find the average value of a function over a region in the plane? the average value of a function of two variables is defined to be one over the area of that region times the double integral over that region of f of x, y, da.
00:48 So this time i have a square that's 2 by 2, and so the area of this region is 4.
00:57 This is times the double integral of our function, so the quantity of x squared plus y squared.
01:08 And since my region is a square, i don't think it matters if i set this up in the dx, d, y, d, y, d, d, y, direction.
01:19 And so for the sake of argument, let's set this up d, y, d x.
01:25 I'll have my y's range from zero to two, and my x is.
01:31 May range from 0 to 2 as well.
01:37 So now, evaluating this double integral, i'm first going to need an anti -derivative with respect to y.
01:47 My anti -derivative of x squared with respect to y is x -squared y.
01:54 My anti -derivative of y squared with respect to y, y -cubed over 3.
02:03 And this is evaluated from 0 to 2, dx.
02:13 And so now we'll evaluate.
02:14 We'll say, okay, i've got a quarter times the integral from zero to two...
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