Question

Find the average value of f over the given rectangle: f(x, y) = 3x^2y, R has vertices (-3, 0), (-3, 3), (3, 3), (3, 0).

Answer 1

Final answer:

the average value of
`\(f(x, y) = 3x^2y\)`over the given rectangle
`the average value of \(f(x, y) = 3x^2y\) over the given rectangle \(R\) is \( (1)/(2) \).`

Step-by-step explanation:

To find the average value of the function
`\(f(x, y) = 3x^2y\) over the given rectangle \(R\), you can use the following formula:`

`\[ \text{Average value} = \frac{1}{\text{Area of } R} \iint_R f(x, y) \, dA \]`

`where \(dA\) is the area element in the plane.`

`Given that \(R\) has vertices \((-3, 0)\), \((-3, 3)\), \((3, 3)\), and \((3, 0)\), the width of the rectangle (\(w\)) is \(3 - (-3) = 6\) units, and the height of the rectangle (\(h\)) is \(3 - 0 = 3\) units.`

`Therefore, the area of \(R\) is \(A = w * h = 6 * 3 = 18\) square units.`

Now, you can set up the double integral to find the average value:

`\[ \text{Average value} = (1)/(18) \iint_R 3x^2y \, dA \]`

To integrate over \(R\), you need to define the limits of integration. In this case, \(x\) ranges from \(-3\) to \(3\) and \(y\) ranges from \(0\) to \(3\).

`\[ \text{Average value} = (1)/(18) \int_(-3)^(3) \int_(0)^(3) 3x^2y \, dy \, dx \]`

Now, perform the integration:

`\[ \text{Average value} = (1)/(18) \int_(-3)^(3) \left[(3)/(2)x^2y^2\right]_(0)^(3) \, dx \]`

`\[ \text{Average value} = (1)/(18) \int_(-3)^(3) (27)/(2)x^2 \, dx \]`

`\[ \text{Average value} = (1)/(18) \left[(9)/(2)x^3\right]_(-3)^(3) \]`

`\[ \text{Average value} = (1)/(18) \left[(9)/(2)(27) - (9)/(2)(-27)\right] \]`

`\[ \text{Average value} = (1)/(18) * (9)/(2) * 54 \]`

`\[ \text{Average value} = (1)/(2) \]`

So, the average value of
`\(f(x, y) = 3x^2y\) over the given rectangle \(R\) is \( (1)/(2) \).`