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Find the average value of f(x,y)=sin(x + y) over

the rectangle 0 < x < π/6, 0 < y < 3π/2

User GokulnathP
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1 Answer

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The average value of f(x, y) = sin(x + y) over the rectangle R is -2/3π.

To find the average value of f(x, y) = sin(x + y) over the rectangle R = [0, π/6] × [0, 3π/2], we can use the formula for the average value of a function over a rectangle:

∫∫_R f(x, y) dA / |R|

where |R| is the area of the rectangle.

In this case, we have:

∫∫_R sin(x + y) dA / |R| = ∫₀^(π/6) ∫₀^(3π/2) sin(x + y) dx dy / [(π/6)(3π/2)]

To evaluate this double integral, we can use double integration by parts. First, we integrate with respect to x:

∫₀^(π/6) ∫₀^(3π/2) sin(x + y) dx dy = ∫₀^(π/6) [-cos(x + y)] ₀^(π/6) dy = ∫₀^(3π/2) cos(y) dy

Next, we integrate with respect to y:

∫₀^(π/6) cos(y) dy = sin(y) ₀^(3π/2) = -1

Therefore, the average value of f(x, y) = sin(x + y) over the rectangle R is:

∫∫_R sin(x + y) dA / |R| = -1 / [(π/6)(3π/2)] = -2/3π

User Mlevit
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7.4k points