Question

State the double integral definition of the area of a surface S given by \(z=f(x, y)\) over the region \(R\) in the \(x y\) -plane.

Short answer

The double integral definition of the area of a surface \(S\) represented by \(z = f(x, y)\) over the region \(R\) in the \(xy\)-plane is given by \[\iint_R \sqrt{1 + \left(f_x (x, y)\right)^2 + \left(f_y (x, y)\right)^2} dA\]

Step-by-step solution

Understand the Problem

The problem statement asks for the double integral definition of the area of a surface S represented by \(z = f(x, y)\) over a region \(R\) in the \(xy\)-plane. This involves the understanding of multivariable calculus, particularly double integrals.

Define the Surface Area Using Double Integral

To define the surface area using a double integral, we have to consider a small element, \(dS\), of the surface \(S\), with \(z = f(x, y)\). If we project \(dS\) vertically onto the \(xy\)-plane, it covers an area of \(dA = dx \cdot dy\) in the \(xy\)-plane. Applying the Pythagorean theorem to the change in \(z\) fraction, we get the surface element \(dS\) is equal to \(\sqrt{1 + \left(\frac{{df}}{{dx}}\right)^2 + \left(\frac{{df}}{{dy}}\right)^2} dA = \sqrt{1 + \left(f_x (x, y)\right)^2 + \left(f_y (x, y)\right)^2} dA\). Therefore, the total surface area of \(S\) over the region \(R\) is given by the double integral: \[\iint_R \sqrt{1 + \left(f_x (x, y)\right)^2 + \left(f_y (x, y)\right)^2} dA\], where \(f_x(x, y)\) and \(f_y(x, y)\) represent the partial derivatives of \(f(x, y)\) in respect to \(x\) and \(y\) respectively.