Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) \(f(x, y)=9-x^{2}\) \(R:\) square with vertices (0,0),(3,0),(0,3),(3,3)

Short answer

The surface area is given by the integral \(\int \int_R \sqrt{1+(2x)^2} \, dx \, dy\) where \(R\) is a square with vertices (0,0), (3,0), (0,3), and (3,3). The exact result depends on carrying out this double integral.

Step-by-step solution

Identifying the region of integration

We're given that \(R\) is the square with vertices at (0,0), (3,0), (0,3), and (3,3). This means we will be integrating over \(0 \leq x \leq 3\) and \(0 \leq y \leq 3\).

Computing the gradient of \(f\)

The function \(f(x,y) = 9-x^2\). To find its gradient, we take its partial derivatives with respect to \(x\) and \(y\). The partial derivative of \(f\) with respect to x is \(-2x\), and since \(f\) does not contain \(y\), the partial derivative of \(f\) with respect to \(y\) is 0. So, the gradient of \(f\), denoted as \(\nabla f\), is \(-2x\).

Calculating the magnitude of the gradient

The magnitude of the gradient of a function is the square root of the sum of the squares of its partial derivatives. In our case, that would be \(\sqrt{(-2x)^2 + 0} = \sqrt{4x^2} = 2|x|\). Since we are working over the region where \(0 \leq x \leq 3\), \(2|x|\) is equal to \(2x\).

Integrating over \(R\)

Finally, we compute the surface area over the region \(R\), which is given by the double integral of the square root of \(1 + (\nabla f)^2\) dA. This simplifies to integrating over the region \(R=\) square with vertices (0,0),(3,0),(0,3),(3,3) using the formula \(\int \int_R \sqrt{1+(2x)^2} \, dx \, dy\). Completing this double integral gives the surface area of the function over \(R\).