Question

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) $$ \begin{array}{l} f(x, y)=\sqrt{x^{2}+y^{2}} \\ R=\\{(x, y): 0 \leq f(x, y) \leq 1\\} \end{array} $$

Short answer

The area of the surface defined by \(z = \sqrt{x^{2}+y^{2}}\) over the region \(R = \{(x, y): 0 \leq f(x, y) \leq 1\}\) is \(2π/3\).

Step-by-step solution

Express the Region R in Polar Coordinates

Since \(f(x, y)=\sqrt{x^{2}+y^{2}}\) and the region \(R\) is defined by \(0 \leq f(x, y) \leq 1\), it is easier to express the region in polar coordinates where \(x = rcos(θ)\) and \(y = rsin(θ)\). Thus, \(f(r,θ)=r\) and the region \(R\) is defined by \(0 \leq r \leq 1\), for \(0 \leq θ \leq 2π\).

Find the Differential of Surface Area

The differential of area \(dA\), in polar coordinates, is given by \(rdrdθ\). So, we will use this to calculate the total surface area.

Compute the Surface Integral

To calculate the total surface area, we compute the double integral of \(f(r,θ)\) with respect to the area \(dA\) over the region \(R\). This gives us the integral \(\int_0^{2π}\int_0^1 r . r dr dθ\) which simplifies to \(\int_0^{2π}\int_0^1 r^2 dr dθ\).

Evaluate the Integral

The double integral can be calculated by first integrating with respect to \(r\), yielding \([r^3/3]_0^1\), and then with respect to \(θ\), yielding \([θ]_0^{2π}\). This evaluates to \(2π/3\).