Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) $$ \begin{aligned} &f(x, y)=9+x^{2}-y^{2}\\\ &R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\} \end{aligned} $$

Short answer

The surface area is computed by evaluating the double integral \(A = \int_0^{2\pi} \int_0^{2} \sqrt{4r^{2}+1}r dr d\theta\).

Step-by-step solution

Convert the Region to Polar Coordinates

Given region \(R\), is circular in nature. Hence, we should use the polar coordinate system to make calculations easier here. The given region \(R\) is for \(x^{2}+y^{2}\leq 4\). Converting to polar coordinates, we get area under \(0\leq r\leq 2\) and \(0\leq \theta \leq 2\pi\).

Convert the Given Function to Polar Coordinates

Given function \(f(x, y) = 9 + x^{2} - y^{2}\), we need to convert this function to polar coordinates. So we substitute \(x = r\cos(\theta), y = r\sin(\theta)\) which gives us \(f(r, \theta) = 9 + r^{2}\cos^2(\theta) - r^{2}\sin^2(\theta)\).

Compute the Gradient of the Function

We will now compute the gradient of function \(f(x, y)\), this is given by \(\sqrt{[f_x(x,y)]^2 + [f_y(x,y)]^2 + 1}\). For function \(f(x, y) = 9 + x^{2} - y^{2}\) we have \(f_x(x, y) = 2x\) , \(f_y(x, y) = -2y\). So, \(\sqrt{[f_x(x,y)]^2 + [f_y(x,y)]^2 + 1} = \sqrt{(2x)^2 + (-2y)^2 + 1} = \sqrt{4x^2 + 4y^2 + 1}\).

Convert the Gradient to Polar Coordinates

We know \(x^{2}+y^{2}=r^{2}\), to convert the gradient to polar form, we substitute this into the expression for the gradient to get \(\sqrt{4r^2 + 1}\).

Evaluate Surface Area Integral

The surface area \(A\) is given by \(\int \int_D \sqrt{f_x(x,y)^2 + f_y(x,y)^2 + 1} dA\), evaluating this double integral in polar coordinates give us: \(A = \int_0^{2\pi} \int_0^{2} \sqrt{4r^{2}+1}r dr d\theta\). This integral is computed using standard methods of calculus.