Question

Find The Average Value Of F(X,Y)=Sin(X+Y) Over

A. The Rectangle 0≤X≤Π/6,0≤Y≤7π/6
B. The Rectangle 0≤X≤3π/2,0≤Y≤5π/4

Answer 1

Evaluating the integral, we get:

`\[(1)/(A)\iint_R f(x,y) \, dA \approx -0.155\]`

Therefore, the average value of
`\(f(x,y) = \sin(x+y)\)` over the given rectangles is approximately:

a) 0.267

b) -0.155

Explanation:

To find the average value of
`\(f(x,y) = \sin(x+y)\)` over the given rectangles, we need to calculate the double integral of
`\(f(x,y)\)` over each rectangle and divide by the area of the rectangle.

a) For the rectangle
`\(0 \leq x \leq (\pi)/(6)\) and \(0 \leq y \leq (7\pi)/(6)\)`, the average value of
`\(f(x,y)\)` is:

`\[(1)/(A)\iint_R f(x,y) \, dA = (1)/((\pi)/(6) \cdot (7\pi)/(6)) \iint_R \sin(x+y) \, dA\]`

Using the change of variables
`\(u = x+y\)` and
`\(v = y\)`, we can transform the integral to the region
`\(0 \leq u \leq (4\pi)/(3)\) and \(0 \leq v \leq (\pi)/(6)\):`

`\[(1)/(A)\iint_R f(x,y) \, dA = (1)/((\pi^2)/(36)) \int_0^{(\pi)/(6)} \int_0^{(4\pi)/(3)-v} \sin(u) \, du \, dv\]`

Evaluating the integral, we get:

`\[(1)/(A)\iint_R f(x,y) \, dA \approx 0.267\]`

b) For the rectangle
`\(0 \leq x \leq (3\pi)/(2)\) and \(0 \leq y \leq (5\pi)/(4)\), the average value of \(f(x,y)\) is:`

`\[(1)/(A)\iint_R f(x,y) \, dA = (1)/((3\pi)/(2) \cdot (5\pi)/(4)) \iint_R \sin(x+y) \, dA\]`

Using the change of variables
`\(u = x+y\) and \(v = y\),` we can transform the integral to the region
`\(0 \leq u \leq (11\pi)/(4)\) and \(0 \leq v \leq (3\pi)/(2)\):`

`\[(1)/(A)\iint_R f(x,y) \, dA = (1)/((15\pi^2)/(8)) \int_0^{(3\pi)/(2)} \int_0^{(11\pi)/(4)-v} \sin(u) \, du \, dv\]`

Evaluating the integral, we get:

`\[(1)/(A)\iint_R f(x,y) \, dA \approx -0.155\]`

Therefore, the average value of
`\(f(x,y) = \sin(x+y)\)` over the given rectangles is approximately:

a) 0.267

b) -0.155